Mika has a rectangular fish tank that is 65 cm wide and 85 cm long. When completely full, the tank holds 221 L of water. She plans to fill the tank? full, and she wants to find the height of the water. 4 1 L - 1000 cm3 volume = xwxh Mika is calculating what the height of the water will be. Choose ALL correct steps that would be included in her calculation Find the height of the tank: 4 A) x 40 = 30 cm 4 Find 3 the height of the tank: 4 4 X 30 - 40 cm Find the height of the tank 3 x 85 - 30 cm 4 DS Divide the length and width by the volume to find height: 65 x 70 - 40 cm 168 x 1000 Divide the volume by the length and width to find height: 221 x 1000 - 40 cm 65 XSS​

To find the total volume of the silo, we need to add the volume of the hemisphere on top to the volume of the cylinder at the bottom.

The volume of a hemisphere is given by:

V_hemi = (2/3)πr^3

where r is the radius of the hemisphere.

Substituting r = 9ft, we get:

V_hemi = (2/3)π(9ft)^3

= 1521π ft^3

The volume of a cylinder is given by:

V_cyl = πr^2h

where r is the radius of the cylinder and h is its height.

Substituting r = 9ft and h = 35ft, we get:

V_cyl = π(9ft)^2(35ft)

= 2673π ft^3

Therefore, the total volume of the silo is:

V_silo = V_hemi + V_cyl

= 1521π + 2673π

= 4194π ft^3

≈ 13160 ft^3

Rounding to the nearest cubic foot, the silo can hold approximately 13160 cubic feet of grain.

I have gotten 113

Step-by-step explanation:

The right angle triangle has three angles and the value of its two angles are 90°and30°.We need to find the third angle which is the sum of 90°and 30°subtract from 180°=60°.60° is vertically opposite to angle C. We'll be having a quadrilateral whose angles add up to 360° .Subtract the sum of the three angles from 360° and you'll get 113°

Answer:

72.5 square inches

Step-by-step explanation:

See attachment.

The areas of the 2 shapes are in blue, but when added together:

60+12.5=72.5

Hope this helps!

Yes, there is a way for this equation to be true in now, take the square root of both sides of the equation (c^2 = n^2) and write the resulting equation

To take the square root of both sides of the equation (c^2 = n^2), you would perform the following steps:

1. Take the square root of both sides:
√(c^2) = √(n^2)

2. Simplify the square roots:
c = n

The resulting equation is c = n.

This equation can be true if both c and n have the same value. This means that c and n could be positive or negative, but their magnitudes must be the same.

For example, if c = 3 and n = 3, then the equation holds true, as both sides are equal.

Similarly, if n = -5, then c could be either 5 or -5, since both values have a magnitude of 5.

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Therefore, the maximum value of f(x,y) inside the triangle is 80/9, which occurs along the line y = (-1/2)x + 4 at the point (8/3, 10/3), and the minimum value is -32, which occurs at the critical point (-8,4).

To find the maximum and minimum values of f(x,y) = xy on the region inside the triangle whose vertices are (6,2), (0,3), and (6,0), we use the method of Lagrange multipliers.

First, we need to find the critical points of f(x,y) subject to the constraint that (x,y) lies inside the triangle. We can express this constraint using the equations of the lines that form the sides of the triangle:

y = (-1/2)x + 4

y = (3/2)x

y = 0

Next, we set up the Lagrange multiplier equation:

∇f = λ∇g

where g(x,y) is the equation of the constraint, i.e., the triangle.

We have:

f(x,y) = xy

∇f = <y, x>

g(x,y) = y - (-1/2)x - 4 = 0

∇g = <-1/2, 1>

Setting ∇f = λ∇g, we get:

y = (-1/2)λ

x = λ

Substituting these into the constraint equation, we get:

(-1/2)λ - 4 = 0

Solving for λ, we get:

λ = -8

Substituting this into y = (-1/2)λ and x = λ, we get:

x = -8 and y = 4

Therefore, the only critical point of f(x,y) inside the triangle is (-8,4).

Next, we need to check the values of f(x,y) at the vertices and along the sides of the triangle.

At the vertices:

f(6,2) = 12

f(0,3) = 0

f(6,0) = 0

Along the line y = (3/2)x:

f(x, (3/2)x) = (3/2)x^2

Using the vertex (6,2) and the x-intercept (4/3, 2), we can see that the maximum value of (3/2)x^2 on this line occurs at x = 4. Therefore, the maximum value of f(x,y) along this line is:

f(4,6) = 24

Along the line y = (-1/2)x + 4:

f(x, (-1/2)x + 4) = (-1/2)x^2 + 4x

Using the vertex (6,2) and the x-intercept (8,0), we can see that the maximum value of (-1/2)x^2 + 4x on this line occurs at x = 8/3. Therefore, the maximum value of f(x,y) along this line is:

f(8/3,10/3) = 80/9

Finally, we need to check the values of f(x,y) at the critical point (-8,4). We have:

f(-8,4) = -32

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The depth of trail mix box that is shaped like a rectangular prism must be 2.4 inches.

We are going to use the formula for determining volume to work out the depth of the rectangular prism.

The volume of a rectangular prism is given by the formula:

V = l × w × h

where:

V = the volume

l = the length

w = the width

h = the height.

Given:

w = 7.5 inches

h = 9 inches

V = 162 cubic inches

Now, we are to find the value of d, the depth of the box, which corresponds to the length of the rectangular prism.

Substituting the values into the formula for volume:

162 = d × 7.5 × 9

Simplifying the right-hand side:

162 = 67.5d

Dividing both sides by 67.5, we get:

d = 162 ÷ 67.5

d = 2.4 inches

Therefore, the depth of the box shaped as rectangular prism = 2.4 inches.

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Step-by-step explanation:

the answer will be "15" according to the question.

The red car is actually traveling along the road at a speed of approximately 26.67 feet per second.

We can start by drawing a diagram of the situation:

     P (police car)

      |

      |

      |

40    |    S (stop sign)

-------|--------------------

      | 160

      |    R (red car)

Let's use the Pythagorean theorem to find the distance between the police car and the red car at any time t:

d(t)² = 40² + (160 - v*t)²

Where v is the speed of the red car in feet per second, and d(t) is the distance between the police car and the red car at time t.

We want to find how fast the red car is actually traveling along the road, so we need to find v when the distance between the police car and the red car is decreasing at a rate of 100 feet per second:

d'(t) = -100

We can take the derivative of the equation for d(t) with respect to time:

2d(t)d'(t) = 0 + 2(160 - v*t)(-v)

Simplifying and plugging in d'(t) = -100, we get:

-4000 + 2v²t = -100(160 - vt)

Solving for v, we get:

v = 80/3 ≈ 26.67 feet per second

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The approximate values of y at x = 0.1, 0.2, 0.3, and 0.4 are all equal to y1 = y2 = y3 = y4 = 0.5, as we only used the first step of Euler's method.

We can use Euler's method with a step size of 0.5 to approximate the solution of the given initial-value problem as follows:

First, we need to find the slope at the initial point (0, -3):

y' = 1 - 2x - 2y

y'(0, -3) = 1 - 2(0) - 2(-3) = 7

Using Euler's method, we can approximate the solution at x = 0.5:

y(0.5) ≈ y(0) + hy'(0, -3) = -3 + 0.57 = 0.5

Next, we can use the approximate value y(0.5) to approximate the solution at x = 1:

y(1) ≈ y(0.5) + hy'(0.5, 0.5) = 0.5 + 0.5(1 - 2(0.5) - 2(0.5)) = -0.5

Similarly, we can use the approximate value y(1) to approximate the solution at x = 1.5:

y(1.5) ≈ y(1) + hy'(1, -0.5) = -0.5 + 0.5(1 - 2(1) - 2(-0.5)) = -1.25

Finally, we can use the approximate value y(1.5) to approximate the solution at x = 2:

y(2) ≈ y(1.5) + hy'(1.5, -1.25) = -1.25 + 0.5(1 - 2(1.5) - 2(-1.25)) = -2.4375

Therefore, the approximate values of y at x = 0.1, 0.2, 0.3, and 0.4 are all equal to y1 = y2 = y3 = y4 = 0.5, as we only used the first step of Euler's method.

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The derivative of the vector function is: r'(t) = (-2t/(7-t^2)) i + (1/(2sqrt(13+t))) j - 36e^(9t) k

We are given a vector function r(t) = ln(7-t^2)i + sqrt(13+t)j – 4e^(9t)k, and we need to find its derivative r'(t).

The derivative of a vector function is obtained by differentiating each component of the vector function separately.

So, let's differentiate each component:

r(t) = ln(7-t^2)i + sqrt(13+t)j – 4e^(9t)k

r'(t) = (d/dt) ln(7-t^2) i + (d/dt) sqrt(13+t) j - (d/dt) 4e^(9t) k

Using the chain rule of differentiation, we have:

r'(t) = -2t/(7-t^2) i + 1/(2sqrt(13+t)) j - 36e^(9t) k

Therefore, the derivative of the vector function is:

r'(t) = (-2t/(7-t^2)) i + (1/(2sqrt(13+t))) j - 36e^(9t) k

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Answer: hopefully the image helps

Step-by-step explanation:

The unique solution, y(x), to the given differential equation with initial condition y(0) = 4 is

:
y(x) = [-(6x⁸ - 64)]¹/³

Determine the unique solution?

To determine the unique solution, y(x), to the given differential equation with initial condition y(0) = 4, we first need to separate the variables and integrate both sides with respect to x and y, respectively.

dy/y⁴ = 8x⁷ dx

Integrating both sides, we get:

-1/3y³ = 2x⁸ + C

where C is the constant of integration.

Now we can use the initial condition y(0) = 4 to solve for C:

-1/3(4)³ = 2(0)⁸ + C

C = -64/3

Substituting C back into the previous equation, we get:

-1/3y³ = 2x⁸ - 64/3

Multiplying both sides by -3 and taking the cube root, we get:

y(x) = [-(6x⁸ - 64)]¹/³

Therefore, the unique solution, y(x), to the given differential equation with initial condition y(0) = 4 is:

y(x) = [-(6x⁸ - 64)]¹/³

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The difference between the abscissa of P and Q is 1.

The abscissa of a point is its x-coordinate, or horizontal distance from the origin (usually measured along the x-axis).

In the given problem, the abscissa of point P is -2, which means it is located 2 units to the left of the origin on the x-axis. The abscissa of point Q is -3, which means it is located 3 units to the left of the origin on the x-axis.

To find the difference between the abscissas of P and Q, we simply subtract the abscissa of Q from the abscissa of P:

(abscissa of P) - (abscissa of Q) = (-2) - (-3) = -2 + 3 = 1

Therefore, the difference between the abscissas of P and Q is 1 unit.

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Answer:

1st choice:  1/4(y - 10) = 2/3

Step-by-step explanation:

the "variable" is y

"is" means "=" (equals sign)

one fourth =  1/4

"difference of" means subtract

Answer:  1/4(y - 10) = 2/3

A function is a rule that assingns each value of independent variable to exactly one value of the dependent variable.

A function is a mathematical concept that relates two sets of values, known as the domain and the range. The domain is the set of independent variables, while the range is the set of dependent variables. A function is a rule that assigns to each value in the domain exactly one value in the range.

For example, if we have a function f(x) = 2x + 3, the domain would be any possible value of x, and the range would be any possible value of 2x + 3. So if we put x = 2, then f(x) = 2(2) + 3 = 7. Therefore, the function assigns the value of 7 to the value of 2 in the domain.

Functions are used in various branches of mathematics, science, and engineering to model and analyze relationships between two or more variables. They are an important concept in calculus, where they are used to study rates of change and optimization problems.

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The integral of f(x, y) = x + y over the given curve is 8.

To integrate the function f(x, y) = x + y over the curve C: x² + y² = 4 in the first quadrant from (2, 0) to (0, 2), we will use the line integral. Since the curve is a circle, we can parameterize it using polar coordinates as follows:

x = 2cos(θ)
y = 2sin(θ)

Now, let's find the derivatives:

dx/dθ = -2sin(θ)
dy/dθ = 2cos(θ)

Next, we substitute x and y in f(x, y):

f(x, y) = 2cos(θ) + 2sin(θ)

Now, we can set up the line integral:

∫[f(x, y) * ||dr/dθ||]dθ

Since ||dr/dθ|| = sqrt((-2sin(θ))^2 + (2cos(θ))^2) = 2, the line integral becomes:

∫[2cos(θ) + 2sin(θ)] * 2 dθ

To find the limits of integration, we can use the points (2, 0) and (0, 2). In polar coordinates, these points correspond to θ = 0 and θ = π/2.

So, the line integral becomes:

∫[4cos(θ) + 4sin(θ)]dθ from 0 to π/2

Now, we can integrate and evaluate:

[4sin(θ) - 4cos(θ)] from 0 to π/2 = [4(1) - 4(0)] - [4(0) - 4(1)] = 8

Thus, the integral of f(x, y) = x + y over the given curve is 8.

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Hello beautiful people! I hope you're all doing well today. As for me, I'm feeling great and ready for the weekend! I plan on spending some quality time with my loved ones, exploring new places, and trying out new activities.

I believe that weekends are meant for rest, relaxation, and rejuvenation, so I'm looking forward to taking a break from my busy work schedule.

One of the things I love about weekends is the opportunity to disconnect from the stresses of everyday life and focus on things that bring me joy. Whether it's going for a hike, trying out a new recipe, or catching up with friends over a cup of coffee, there's always something to look forward to.

this weekend I plan on making the most of my free time by doing things that make me happy and help me recharge. I hope you all have a great weekend as well, and that you find time to do the things you love with the people you care about. Remember, life is short, so let's make the most of every moment we have!

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The value of v in the equation is 26

The equation is given as

v= u + at

The parameters given are

u= 6

a= 10

t= 2

v= 6 + 10(2)

v= 6 + 20

v= 26

Hence the value of v in the equation is 26

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The correct number of beads is; 4 black, 8 red, and 4 blue. Option A

A molecular model is a depiction of molecules or chemical compounds made physically, visually, or mathematically in order to comprehend their behavior and characteristics. These models can range from real models made of plastic or metal to computer-generated graphics or mathematical formulae, and they can be straightforward or sophisticated.

There are four carbon atoms, eight hydrogen atoms and four oxygen atoms.

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60 ≤ 1.36x + 34 ≤ 85; To stay within the range, the usage should be between 19.1 and 37.5 HCF.

The correct compound inequality to represent the scenario is:

60 ≤ 1.36x + 34 ≤ 85

To solve for x, we need to isolate it in the middle of the inequality:

60 - 34 ≤ 1.36x ≤ 85 - 34

26 ≤ 1.36x ≤ 51

Finally, we divide by 1.36 to isolate x:

19.12 ≤ x ≤ 37.5

Therefore, the recommended range of water consumption is between 19.1 and 37.5 HCF. The answer is (D) 60 ≤ 1.36x + 34 ≤ 85; To stay within the range, the usage should be between 19.1 and 37.5 HCF.

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complete question

To conserve water, many communities have developed water restrictions. The water utility charges a fee of $34, plus an additional $1.36 per hundred cubic feet (HCF) of water. The recommended monthly bill for a household is between $60 and $85 dollars per month. If x represents the water usage in HCF in a household, write a compound inequality to represent the scenario and then determine the recommended range of water consumption. (Round your answer to one decimal place.)

60 ≤ 1.36x − 34 ≤ 85; To stay within the range, the usage should be between 69.1 and 87.5 HCF.

60 ≤ 1.36x − 34 ≤ 85; To stay within the range, the usage should be between 44.1 and 87.5 HCF.

60 ≤ 1.36x + 34 ≤ 85; To stay within the range, the usage should be between 37.5 and 44.1 HCF.

60 ≤ 1.36x + 34 ≤ 85; To stay within the range, the usage should be between 19.1 and 37.5 HCF.

Determine whether the two figures are similar: D. No, the two figures are not similar.

In Geometry, two (2) quadrilaterals are similar when the ratio of their corresponding sides are equal in magnitude and their corresponding angles are congruent.

Additionally, two (2) geometric figures such as quadrilaterals are considered to be congruent only when their corresponding side lengths are congruent (proportional) and the magnitude of their angles are congruent;

Ratio = 12/8 = 10/6 = 10/6

Ratio = 3/2 ≠ 5/3 = 5/3

In conclusion, the two figures are not similar because the ratio of their corresponding sides is not proportional.

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

The function that models this situation is f(n) = -3n + 24.

To find the function, we need to analyze the relationship between the number of meals dispensed (n) and the amount of pet food remaining (f(n)).

1. Observe the change in f(n) when n increases by 1 meal. From n=1 to n=3, f(n) decreases from 21 to 15, a change of -6. From n=6 to n=7, f(n) decreases from 6 to 3, a change of -3.
2. The decrease in f(n) is not constant, so the function is not linear. However, the decrease becomes smaller as n increases.
3. Consider the average rate of change in f(n) per meal: (-6/2) = -3, (-3/1) = -3.
4. Since the average rate of change is constant, the function is linear.
5. The function has the form f(n) = -3n + b. To find b, plug in the value of n and f(n) from the table: 21 = -3(1) + b, which gives b = 24.
6. Therefore, the function that models this situation is f(n) = -3n + 24.

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Answer:

? equals 80

Step-by-step explanation:

After 2 hours, approximately 4.38 grams of Francium remain from the 480-gram sample.

What is Algebraic expression ?

An algebraic expression is a combination of variables, constants, and mathematical operations such as addition, subtraction, multiplication, and division. It may contain one or more terms, with each term separated by a plus or minus sign. Algebraic expressions are used in algebra to represent mathematical relationships and formulas.

a) To write an exponential function that models the mass of Francium remaining after t minutes, we can use the formula:

N = N0 * [tex](1/2)^{(t / t1/2)}[/tex]

where N is the amount remaining after time t, N0 is the initial amount, t1/2 is the half-life, and (t/t1/2) means raised to the power of t/t1/2.

In this case, the initial amount is 480 grams, the half-life is 22 minutes, and we want to find the amount remaining after t minutes. Therefore, the exponential function that models the mass of Francium remaining after t minutes is:

N = 480 * [tex](1/2)^{t/22}[/tex]

b) 2 hours is equal to 120 minutes. To find how many grams of Francium remain after 2 hours, we can substitute t = 120 into the exponential function we found in part a):

N = 480 *[tex](1/2)^{ (120 / 22) }[/tex] ≈ 4.38 grams

Therefore, after 2 hours, approximately 4.38 grams of Francium remain from the 480-gram sample.

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If c represents the cost per pound for the plums, the cost per pound for the plums is $0.90.

First, we need to determine the total cost of the plums before the coupon was applied.

4 1/5 pounds can be written as a mixed number:

4 1/5 = 21/5

So, the total cost of the plums without the coupon can be found by multiplying the cost per pound (c) by 21/5:

Total cost = c * 21/5

Now we can create an equation to represent the total cost after the coupon was applied:

Total cost - coupon = $3.23

Substituting the expression for total cost:

c * 21/5 - 0.55 = 3.23

To solve for c, we can start by adding 0.55 to both sides:

c * 21/5 = 3.78

Then, we can isolate c by multiplying both sides by the reciprocal of 21/5:

c = 3.78 / (21/5)

c = 0.90

Therefore, the cost per pound for the plums is $0.90.

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There are 42 yellow skittles in the box.

Based on the given information, we know that the ratio of green skittles to yellow skittles is 5:7. This means that for every 5 green skittles, there are 7 yellow skittles.

To find out how many yellow skittles are in the box, we need to know how many sets of 5 green skittles there are. We can do this by dividing the total number of skittles in the box (30) by 5 (since there are 5 green skittles for every set).
30 ÷ 5 = 6

This means there are 6 sets of 5 green skittles in the box.

Now we can use the ratio of 5:7 to find out how many yellow skittles there are in each set:
5 green skittles : 7 yellow skittles

Since there are 7 yellow skittles in each set, we can find the total number of yellow skittles by multiplying 7 by the number of sets (6):
7 x 6 = 42

There are 42 yellow skittles in the box.

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The point with the least absolute value on the number line is always the point zero.

The absolute value of a number is the distance that number is from zero on the number line. Therefore, the point on the number line with the least absolute value is the point closest to zero. This point is located at zero itself, as it is the point on the number line that is equidistant from both the positive and negative numbers.

To further explain, consider the following examples:

- The point 3 is 3 units away from zero, but the point -3 is also 3 units away from zero.
- The point 5 is 5 units away from zero, but the point -5 is also 5 units away from zero.
- The point 0 is 0 units away from zero, making it the point with the least absolute value on the number line.

In conclusion, the point with the least absolute value on the number line is always the point zero.

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