Projectile motion. the height, y metres, of a ball after it has been hit can be modelled by the equation y = -1/8x²+x+c, where x is the horizontal distance travelled by the object in metres and c is a constant. (i) given that the maximum height attained by the ball is 2.4 metres, find the value of c and the corresponding horizontal distance travelled by the ball.​

Answer:

68in

Step-by-step explanation:

square: 3x3=9

9+9= 18

trapezoid (centre area) : 7x3x5=25

25x2=50

50+18=68

not entirely sure abt this

Answer:

26r^2 + 36r.

Step-by-step explanation:

Answer:

f(0) = 1

g(-2) = 3

f(-7)= und

g(4) x f(3) = -2 x 0 = 0

g(-4) = 2

g(x) = 0 --> x = 6, 0.5

f(x) = -1 --> x = -3, 5

f(g(3)) = f(-3) = -1

g(f(-2) = g(0) = -3

f(g(1)) = f(-3) = -1

f(g(5)) = f(-1) = 1

g(f(-4)) = g(-2) = 2

g(g(-6)) = g(4) = -2

g(f(0)) = g(1) = -3

g(f(-6)) = und

Step-by-step explanation:

In order to find the first group, such as f(0), you want to look at the f graph and find 0 on the x-axis. Wherever the y coordinate is will be the correct answer.

To find one such as f(g(3)), you want to dissect it like it is 2 problems. First, we want to find g(3) which is -3. Then we will find -3 on the f graph and find the answer with that y-coordinate.

The given set of questions are solved under the condition of  parametric equations x(t)=sin(3t) and y(t)=cos(3t) .

Hence, the length of the curve from t= 0 to t= π is 3π.

Now,

A. To evaluate the velocity, we need to perform the derivative of x(t) and y(t) concerning t.

x'(t) = 3cos(3t)

y'(t) = -3sin(3t)

Therefore,  the velocity vector is

v(t) = <3cos(3t), -3sin(3t)>

B. To define the acceleration, we need to evaluate  the derivative of v(t) concerning t.

a(t) = v'(t) = <-9sin(3t), -9cos(3t)>

C. To describe  the speed, we need to calculate  the magnitude of the velocity vector.

|v(t)| = √((3cos(3t))² + (-3sin(3t))²)

= 3

D. In order to find the number of times at which the particle stops, to find when the speed is equal to zero.

|v(t)| = 0 when cos(3t) = 0

sin(3t) = 0.

Therefore,

cos(3t) = 0 when t = (π/6) + (nπ/3),

here n = integer.

sin(3t) = 0 when t = (nπ/3),

here n = integer.

E. To calculate the length of the curve from t=0 to t=π by performing  calculus

L = ∫[a,b] √((dx/dt)² + (dy/dt)²) dt

Therefore, a=0 and b=π.

L = ∫[0,π] √((3cos(3t))² + (-3sin(3t))²) dt

 = ∫[0,π] 3 dt

 = 3π

The given set of questions are solved under the condition of  parametric equations x(t)=sin(3t) and y(t)=cos(3t) .

Hence, the length of the curve from t=0 to t=π is 3π.

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The complete question is

A particle moves along a path in the xy-plane. the path is given by

the parametric equations x(t)=sin(3t) and y(t)=cos(3t), help with

steps A-E

a. Find the velocity

b. Find the acceleration.

c. Find the speed and simplify your answer completely.

d. Find any times at which the particle stops. Thoroughly explain your answer.

e. Use calculus to  find the length of the curve from t=0 to t = π , show your work.

Answer:

Function A: f(x) = -8^x

Function B: f(x) = 2^x

Function A has a greater horizontal asymptote. As x approaches negative infinity, Function A approaches y = 0 faster than Function B.

Answer:

Step-by-step explanation:

The best-supported statement by the information in the table is:

C. The fee at Bank K will always be greater than the fee at Bank F.

This is because for any given average daily balance, Bank K has a higher fee than Bank F. For example, for an average daily balance of less than $500, Bank F charges $8 while Bank K charges $12. Similarly, for an average daily balance between $500 and $750, Bank F charges no fee while Bank K charges $12. Therefore, the fee at Bank K will always be greater than the fee at Bank F.

By the end of 12 months, Claire will need 1,596,018 pens to house all the rabbits.

The number of adult rabbits in each month is the sum of the adult rabbits from the previous month and the number of baby rabbits that have grown to adulthood.

The number of baby rabbits in each month is the product of the number of adult rabbits in the previous month and the number of kits each pair of rabbits produces (6 in this case).

The total number of rabbits is simply the sum of the number of adult rabbits and the number of baby rabbits. Finally, the minimum pen size needed is found by divide the total number of rabbits by 2 (since each rabbit needs 2 square feet of space).

Month 1

Beginning of month: 2 rabbits (1 male, 1 female)

End of month: 8 rabbits (3 males, 5 females)

Number of pens needed: 16 square feet / 2 square feet per pen = 8 pens

Month 2

Beginning of month: 8 rabbits (3 males, 5 females)

End of month: 26 rabbits (11 males, 15 females)

Number of pens needed: 52 square feet / 2 square feet per pen = 26 pens

Month 3

Beginning of month: 26 rabbits (11 males, 15 females)

End of month: 80 rabbits (35 males, 45 females)

Number of pens needed: 160 square feet / 2 square feet per pen = 80 pens

Month 4

Beginning of month: 80 rabbits (35 males, 45 females)

End of month: 242 rabbits (105 males, 137 females)

Number of pens needed: 484 square feet / 2 square feet per pen = 242 pens

Month 5

Beginning of month: 242 rabbits (105 males, 137 females)

End of month: 728 rabbits (315 males, 413 females)

Number of pens needed: 1456 square feet / 2 square feet per pen = 728 pens

Month 6

Beginning of month: 728 rabbits (315 males, 413 females)

End of month: 2186 rabbits (945 males, 1241 females)

Number of pens needed: 4372 square feet / 2 square feet per pen = 2186 pens

Now, to continue for the next 6 months

Month 7

Beginning of month: 2186 rabbits (945 males, 1241 females)

End of month: 6568 rabbits (2835 males, 3733 females)

Number of pens needed: 13136 square feet / 2 square feet per pen = 6568 pens

Month 8

Beginning of month: 6568 rabbits (2835 males, 3733 females)

End of month: 19702 rabbits (8499 males, 11203 females)

Number of pens needed: 39404 square feet / 2 square feet per pen = 19702 pens

Month 9

Beginning of month: 19702 rabbits (8499 males, 11203 females)

End of month: 59110 rabbits (25499 males, 33611 females)

Number of pens needed: 118220 square feet / 2 square feet per pen = 59110 pens

Month 10

Beginning of month: 59110 rabbits (25499 males, 33611 females)

End of month: 177334 rabbits (76535 males, 100799 females)

Number of pens needed: 354668 square feet / 2 square feet per pen = 177334 pens

Month 11

Beginning of month: 177334 rabbits (76535 males, 100799 females)

End of month: 532006 rabbits (229799 males, 302207 females)

Number of pens needed: 1064012 square feet / 2 square feet per pen = 532006 pens

Month 12

Beginning of month: 532006 rabbits (229799 males, 302207 females)

End of month: 1596018 rabbits (689397 males, 906621 females)

Number of pens needed: 3192036 square feet / 2 square feet per pen = 1596018 pens

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In this carnival game, a player flips a coin that has a 3 on one side and a 7 on the other, and then rolls a number cube that has numbers 1-6.

To determine the probabilities, we need to analyze each event separately and then use the multiplication rule of probability to find the probability of both events happening together.

The probability of getting a 3 on the coin is 50%, since there are only two possible outcomes. The probability of rolling each number on the cube is 16.67%, since the cube has six sides.

The probability of both events happening together depends on the individual probabilities and is found by multiplying them. Finally, we can use the addition rule of probability to find the probability of either event happening.

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Answer:

425

Step-by-step explanation:

212,5*2=425

I'm sorry, but I cannot determine the values of the line segment lengths

and angle measures without further information or context about the

geometry problem. Please provide me with more details or the specific

problem in question.

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Answer:

  See attached

Step-by-step explanation:

You want a graph with points plotted at (4, 6) and (7, 2), representing a sand trap and a water hazard, respectively.

The ordered pair (4, 6) represents the coordinates (x, y). The x-coordinate is the number of units right of the point x=0, and the y-coordinate is the number of units up from y=0.

Both points have positive coordinates for both x and y, so will be located up and right from the origin. The plot is shown in the attachment.

<951414049393>

The equation would be written as: y = $85,000(1 + 0.12/1)^5

Then the predicted earning would be y = $132,559

y=a(b)x

where y = income

a =  $85,000

b = (1 + r = percent increase)

then x = time period = 5 years

When we put in the values we would have y = $85,000(1 + 0.12/1) ^5

The exponential function of the form y = a(b)^x is: y = $85,000(1 + 0.12/1) ^5

When we solve the above, we would have the income = y = $132,559

Therefore the predicted earnings that Dans small business would have in a period of five years is equal to $132,559

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well, according the Cavalieri's Principle, the volume of the oblique cone will be the same volume as the non-oblique cone, so

[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=9\\ h=16 \end{cases}\implies V=\cfrac{\pi (9)^2(16)}{3}\implies V=432\pi ~cm^3[/tex]

The total area of the soccer field is approximately 8150 square yards.

We'll find the total area of the soccer field in square meters first, and then convert it to square yards using the conversion factor provided.

Find the area in square meters (m²):
Area = Length × Width
Area = 102.9 m × 66.3 m
Area ≈ 6816.47 m²

Convert the area to square yards (yd²):
Use the conversion factor: 1 yard = 0.9144 meters
1 m² = (1/0.9144)² yd²
1 m² ≈ 1.19599 yd²
Now, multiply the area in m² by the conversion factor to get the area in yd²:
Area ≈ 6816.47 m² × 1.19599 yd²/m²
Area ≈ 8150 yd² (rounded to the nearest whole number).

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There are 4 possible sets of dimensions for the cuboid with a volume of 1815 cm³.

First, find the prime factors of 1815:

1815 = 3 × 5 × 11 × 11

Now, we need to find all possible combinations of these factors into three whole numbers. Each combination of three numbers, when multiplied, should give 1815. We can do this by finding the different ways the prime factors can be distributed among the three dimensions:

3 × 5 × (11 × 11) = 15 × 121 (height × width × length)

3 × 11 × (5 × 11) = 33 × 55

5 × 11 × (3 × 11) = 55 × 33

11 × 11 × (3 × 5) = 121 × 15

We have found 4 different sets of dimensions for the cuboid:

15 × 121 × 1

33 × 55 × 1

55 × 33 × 1

121 × 15 × 1

There are 4 possible sets of dimensions for the cuboid with a volume of 1815 cm³.

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Answer: H = 5

the equation would be, 35h - 25 = $200

Step-by-step explanation:

Ms. Yamagata needs to buy at least 2628 6-inch tiles to cover the floor of her rectangular bathroom.

To determine the least number of 6-inch tiles Ms. Yamagata needs to buy to cover her 9 feet long and 73 feet wide bathroom floor, follow these steps:

1. Convert the dimensions of the bathroom to inches, as the tiles are measured in inches:
  9 feet * 12 inches/foot = 108 inches long
  73 feet * 12 inches/foot = 876 inches wide

2. Determine the total area of the bathroom in square inches:
  Area = length * width = 108 inches * 876 inches = 94,608 square inches

3. Calculate the area of a single 6-inch tile:
  Area = length * width = 6 inches * 6 inches = 36 square inches

4. Divide the total area of the bathroom by the area of a single tile to find the least number of tiles needed:
  Number of tiles = total area / tile area = 94,608 square inches / 36 square inches ≈ 2,628.56

Since Ms. Yamagata cannot buy a fraction of a tile, she needs to buy at least 2,629 6-inch tiles to cover her bathroom floor.

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The radius of convergence R is 1/4 and the interval of convergence is (-6.25, -5.75) for the power series

∑((-4[tex])^n[/tex]) * (-(x + 6[tex])^n[/tex]) / sqrt(n)
To find the radius of convergence (R) and interval of convergence for the power series ∑((-4[tex])^n[/tex]) * (-(x + 6[tex])^n[/tex]) / sqrt(n)

where n starts from 1 to infinity,

We can use the Ratio Test.
Step 1: Apply the Ratio Test
We want to find the limit as n approaches infinity of the absolute value of the (n+1)th term divided by the nth term:
lim (n→∞) |((-4[tex])^{(n+1)[/tex] * (-(x + 6)^(n+1)) / sqrt(n+1)) / ([tex](-4)^n[/tex] * (-(x + 6[tex])^n[/tex]) / sqrt(n))|
Step 2: Simplify the expression
The limit simplifies to:
lim (n→∞) |((-4)(x + 6))/sqrt((n+1)/n)|
Step 3: Find when the limit is less than 1
For the series to converge, the limit must be less than 1:
|(-4)(x + 6)| / sqrt((n+1)/n) < 1
As n approaches infinity, (n+1)/n approaches 1, so the expression simplifies to:
|-4(x + 6)| < 1
Step 4: Determine the radius of convergence (R)
Divide both sides by 4:
|-(x + 6)| < 1/4
The radius of convergence, R, is 1/4.
Step 5: Determine the interval of convergence
To find the interval of convergence, solve for x:
-1/4 < (x + 6) < 1/4
-1/4 - 6 < x < 1/4 - 6
-6.25 < x < -5.75
Thus, the interval of convergence is (-6.25, -5.75).
In summary, the radius of convergence R is 1/4 and the interval of convergence is (-6.25, -5.75).

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Partial sums are:

a) limx→∞ S = L
b) The limit does not exist.
c) limx→∞ S = L
d) The limit does not exist.

We need to use the formulas for partial sums and limits of sequences.

First, recall that the nth partial sum of a series is given by:

Sn = a1 + a2 + ... + an

And the limit of a sequence (if it exists) is given by:

limn→∞ an

Now, let's use these formulas to answer the parts of the question:

a) Find lim S as n approaches infinity:

We have:

S = ax = a1 + a2 + a3 + ... + ax

Taking the limit as x approaches infinity, we get:

limx→∞ S = limx→∞ (a1 + a2 + a3 + ... + ax) = limn→∞ Sn

But we know that the series is convergent, so the limit of the partial sums exists and is equal to the sum of the series:

limn→∞ Sn = L

Therefore:

limx→∞ S = L

b) Find lim as x approaches 0:

We have:

S = ax = a1 + a2 + a3 + ... + ax

Taking the limit as x approaches 0, we get:

limx→0 S = limx→0 (a1 + a2 + a3 + ... + ax)

But as x approaches 0, the number of terms in the sum approaches infinity, so this limit does not exist.

c) Find lim S as x approaches infinity:

We have:

S = ax = a1 + a2 + a3 + ... + ax

Taking the limit as x approaches infinity, we get:

limx→∞ S = limx→∞ (a1 + a2 + a3 + ... + ax) = limn→∞ Sn

Again, we know that the limit of the partial sums exists and is equal to the sum of the series:

limn→∞ Sn = L

Therefore:

limx→∞ S = L

d) Find lim as x approaches 100:

We have:

S = ax = a1 + a2 + a3 + ... + ax

Taking the limit as x approaches 100, we get:

limx→100 S = limx→100 (a1 + a2 + a3 + ... + ax)

But as x approaches 100, the number of terms in the sum approaches infinity, so this limit does not exist.

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1) Note that t is either 4 or -0.25 by virtue of the quadratic function.

2) the vertex and line of summer try are t = 1.875. See the attached graph.

First, identify the values of a, b, and c in the equation...

a = -16

b = 60

c = 16

substitute these values into the quadratic formula

t = (-b ± √(b² - 4ac)) / 2a

t = (-60 ± √(60² - 4(-16)(16))) / 2(-16)

t = (-60 ± √(3600 + 1024)) / (-32)

t = (-60 ± √(4624)) / (-32)

t = (-60 ± 68) / (-32)

So, t can be:

t = (-60 + 68) / (-32) = -1/4

or

t = (-60 - 68) / (-32) = 4

2) To find the line of symmetry, we used t = -b/2a

-60/2(-16)

t = 1.875

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Considering the function f(x) = x(x-4), if the point (2+c, y) is on the graph of f(x), then the following point will also be on the graph of f(x): (2-c, y). Explanation: Since f(x) is symmetric with respect to the vertical line x = 2 (due to the fact that f(x) = x(x-4) = (x-2+2)(x-2) = (x-2)^2 - 2^2), if the point (2+c, y) is on the graph, then its symmetric counterpart, (2-c, y), will also be on the graph.

The definition of a function in mathematics can also be interpreted as a relation that connects each member of x in a set called the domain with a single value f(x) from a second set called the codomain.

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Tip: When writing math questions, the coefficient would go in front of variables: 7b*[tex]b^2[/tex]

Anyways, in this case, since we cannot mutiply 7 by b, the 7 will stay put

But.... we can mutiply b by b

[tex]b^2[/tex][tex]*b[/tex]=b^3

Tip: B's with no exponent has an exponent with one

Questions?

So,7 will stay put and b and b^2 will simply to b^3

[tex]7b^3[/tex]=Answer

a. d, days since first measurement y, acres burned since fire started

0 0, 1 4800, 2 9600, 3 14400, 4 19200, 5 24000. b. when d = -1, it tells that the area burned in the fire was 4780 acres one day before the area was first measured. When d = -3, y = 4680, this means that the area burned in the fire was 4680 acres three days before the area was first measured. c. The area burned a week before the fire measured 4800 acres was approximately 115.2 acres.

a. To complete the table, we need to plug in the values of d in the given equation and calculate the corresponding values of y.

d, days since first measurement

y, acres burned since fire started

0 0

1 4800

2 9600

3 14400

4 19200

5 24000

b. When d = -1, we have:

y = (4800)(24^(-1))^(1) = 4800 - 20 = 4780

This means that the area burned in the fire was 4780 acres one day before the area was first measured.

When d = -3, we have:

y = (4800)(24^(-3))^(1) = 4800 - 120 = 4680

This means that the area burned in the fire was 4680 acres three days before the area was first measured.

c. A week before the fire measured 4800 acres, the number of days since the fire started would be:

d = 4800 / (4800/24) = 24

Therefore, a week before the fire measured 4800 acres, the fire had been burning for 24 days. Plugging in this value in the given equation, we get:

y = (4800)(24^(-24/24))^(1) = 115.2

So, the area burned a week before the fire measured 4800 acres was approximately 115.2 acres.

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Answer:

≈ 12,78 m^2

Step-by-step explanation:

The surface area is equal to the sum of the areas of all the sides

This figure has sides of 2 triangles (bases) and 3 rectangles (lateral surface)

h (triangle) = 1m

We can find the base of the triangle by using the Pythagorean theorem (multiply by 2, because the triangle's base contains two of these identicals sides)

[tex](( {1.5})^{2} - {1}^{2} ) \times 2=( 2.25 - 1 ) \times 2= 1.25 \times 2 = 2.5> 0[/tex]

The triangle's base is equal to:

[tex] \sqrt{2.5} = \frac{ \sqrt{10} }{2} [/tex]

First, let's find the area of 2 bases (triangles):

[tex]a(bases) = 2 \times \frac{1}{2} \times 1 \times \frac{ \sqrt{10} }{2} = \frac{ \sqrt{10} }{2} [/tex]

Now, we can find the whole surface area by adding the areas of the rectangles to the bases' areas:

[tex]a(surface) = \frac{ \sqrt{10} }{2} + 2.4 \times 2 + 1.5 \times 2 + 1.7 \times 2 = \frac{ \sqrt{10} }{2} + \frac{56}{5} ≈12.78[/tex]

The main difference is that Sierra needs to create two equal arcs, while Keaton needs to create six equal arcs to form the vertices of their respective shapes.

For Sierra's inscribed square:
1. Draw a circle with a compass.
2. Mark a point on the circle as one vertex of the square.
3. Draw a diameter passing through the marked point.
4. Use the compass to create two equal arcs, one on each end of the diameter, intersecting the circle.
5. Connect the intersection points to create the square.

For Keaton's inscribed regular hexagon:
1. Draw a circle with a compass.
2. Mark a point on the circle as one vertex of the hexagon.
3. Use the compass to create six equal arcs around the circle, each arc intersecting the end of the previous arc.
4. Connect the intersection points to create the hexagon.

The main difference is that Sierra needs to create two equal arcs, while Keaton needs to create six equal arcs to form the vertices of their respective shapes.

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To find the slope of the secant line connecting two points on a curve, we can use the formula: slope of secant line = (change in y) / (change in x).

In this problem, we are given the function f(x) = x^3 - 5x on the closed interval [-1, 3), and we need to find the slope of the secant line connecting (-1, f(-1)) and (3, f(3)). To do this, we first need to find the y-coordinates of these points by plugging the given x-values into the function f(x). Then, we can use the formula for the slope of a secant line to find the slope of the line connecting these two points.

The slope of a secant line is an important concept in calculus and is used to approximate the instantaneous rate of change of a function at a particular point.

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Tank B fills faster than Tank A, taking approximately 1.75 minutes to fill while Tank A takes approximately 1.14 minutes.

The problem presents two aquarium tanks, A and B, which are of the same size but have different water levels and fill rates. Tank A has 2 feet of water and is being filled at a rate of 2.2 inches per minute, while Tank B has 8 feet of water and is being filled at a faster rate of 5 inches per minute. The goal is to determine how long it will take to fill each tank.

To solve this problem, we need to use the formula: Time = Volume / Rate. We know that the volume of each tank is the same, so we can set up two equations:

For Tank A: Time = (2 feet * 12 inches/foot) / 2.2 inches/minute = 10.91 minutes or approximately 1.14 minutes.

For Tank B: Time = (8 feet * 12 inches/foot) / 5 inches/minute = 19.2 minutes or approximately 1.75 minutes.

Therefore, Tank A will take approximately 1.14 minutes to fill, while Tank B will take approximately 1.75 minutes to fill. It is important to note that Tank B is being filled at a faster rate than Tank A, despite having a greater volume of water.

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Answer:

the answer is the 1st one

2+2=4

3+1=4

4+0 = 4

The dimensions of the cylinder that minimize the production cost are:

radius = approximately 2.78 cm

height = 600 / (πr^2) ≈ 7.14 cm

paper top radius = 2.5 cm

Let's start by finding the formula for the cost of the container in terms of its dimensions.

The volume of the cylinder is given as 600 cubic centimeters, so we have:

πr^2h = 600

where r is the radius of the cylinder and h is its height. Solving for h, we get:

h = 600 / (πr^2)

The surface area of the cylinder is given by:

A = 2πrh + 2πr^2

Substituting h in terms of r, we get:

A = 2πr(600/(πr^2)) + 2πr^2

= 1200/r + 2πr^2

The cost of the container is the sum of the cost of the styrofoam sides and bottom and the cost of the paper top. Let's call the radius of the paper top R, and assume that the height of the cylinder is greater than or equal to the radius of the paper top, so that the top can be completely covered with paper. Then the cost of the container is:

C = 0.04(2πrh + πr^2) + 0.05(πR^2)

Substituting h in terms of r, we get:

C = 0.08πr(600/(πr^2)) + 0.04πr^2 + 0.05πR^2

= 4.8/r + 0.04πr^2 + 0.05πR^2

To minimize the cost, we need to find the values of r and R that minimize the cost function C. To do this, we take the partial derivatives of C with respect to r and R, and set them equal to zero:

dC/dr = -4.8/r^2 + 0.08πr = 0

dC/dR = 0.1πR = 0

Solving for r and R, we get:

r = ∛(60/π) ≈ 2.78 cm

R = 2.5 cm

We can check that these values give us a minimum by checking the second derivatives:

d^2C/dr^2 = 9.6/r^3 + 0.08π > 0 (minimum)

d^2C/dR^2 = 0.1π > 0 (minimum)

Therefore, the dimensions of the cylinder that minimize the production cost are:

radius = approximately 2.78 cm

height = 600 / (πr^2) ≈ 7.14 cm

paper top radius = 2.5 cm

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