Find the exact value of each of the remaining trigonometric functions of θ.
tan θ= -3/5, sec θ>0.

The loan will increase to $39599.29 in 5 years

From the question, we have the following parameters that can be used in our computation:

Using the above as a guide, we have the following:

Amount = P * (1 + r)ᵗ

Where

P = Principal = 35000

r = Rate = 2.5%

t = time = 5

Substitute the known values in the above equation, so, we have the following representation

Amount = 35000 * (1 + 2.5%)⁵

Evaluate

Amount = 39599.29

Hence, the amount is 39599.29

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Answer:

1/3

Step-by-step explanation:

There are 15 tables and 5 have a purple centerpiece. Divide 15 by 5 to get the probability of a randomly selected table being purple.

5/15 is 5 out of 15. 5/15 simplifies to 1/3.

Using the quadratic formula to solve the equation x²-3x+ 5 = 0, the resultant answer is (D) 3+√11/2 and 3-√11/2.

The quadratic formula in elementary algebra is a formula that yields the answer to a quadratic problem.

In addition to the quadratic formula, other methods of solving quadratic equations include factoring, completing the square, graphing, and others.

A second-order equation of the form ax² + bx + c = 0 denotes a quadratic equation, where a, b, and c are real number coefficients and a 0.

So, we have the equation:

x²-3x+ 5 = 0

Now, solve it using the quadratic formula as follows:

x²-3x+ 5 = 0

a = 1

b = -3

c = 5

x = -(-3)±√(-3)² -4*1*5/2

Solve this further:

x = 3±√11/2

Therefore, using the quadratic formula to solve the equation x²-3x+ 5 = 0, the resultant answer is (D) 3+√11/2 and 3-√11/2.

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Correct question:

Solve x²-3x+ 5 = 0.

A.3+√-29 +VE/2 and 3- 2-29

B. 3+√29 and 3-√29

C. 3+√-11 and 3-√-11/2

D. 3+√11/2 and 3-√11/2

Using the table feature of a graphing utility  the intensity of an earthquake with an amplitude of 200,000 micrometers for 0.5 seconds would be 2.16 x 10¹⁴ joules/second/meter².

First, note that the   intensity of an earthquake with an amplitude of 200,000 micrometers for 0.5 seconds can be calculated using the formula..

I = (1/2)ρv²

Here I   Intensity

ρ = density of the medium - this is a constatnt

v = velocity of the S-waves

If we used a graphing tool

a coumun ffor v will be created

a column for the formula for velocity will also be created.

The formula for velocity is

V = A/ T

A = amplitude

T = time or period in seconds


Since A = 200,000 micrometers; and

T = 0.05 secons

v = 200,000 / 0.5

v = 400,000 micrometer/sec

Back o the table, we ahve to add another oclumn for V²

then enter the values for V²

V² = 400,000²
v² = 160,000,000,000 micrometer²/sec²


Nowe we add another column which is

I = 1/2(pv²)

I = 2.16 x 10¹⁴ joules/second/meter².


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Answer:

Distance = Speed × Time.

Let's break down the problem into two parts:

a) At what time will the cyclists meet on the way?

Let's assume t represents the time in hours when the cyclists meet.

The distance Joel travels in t hours is given by: Distance = Speed × Time = 12t km.

The distance Ryan travels in t hours is given by: Distance = Speed × Time = 8t km.

Since they are cycling towards each other, the sum of their distances should equal the total distance between Kingston and Sunny Place, which is 35 km.

So, we can set up the equation: 12t + 8t = 35.

Simplifying the equation, we have: 20t = 35.

Dividing both sides of the equation by 20, we get: t = 35/20 = 1.75 hours.

Therefore, the cyclists will meet on the way after 1.75 hours, which is equivalent to 1 hour and 45 minutes.

b) How far will each cyclist have traveled when they meet?

To find the distance traveled by each cyclist, we can substitute the value of t into their respective equations:

Distance traveled by Joel = 12t = 12 × 1.75 = 21 km.

Distance traveled by Ryan = 8t = 8 × 1.75 = 14 km.

Therefore, when the cyclists meet, Joel would have traveled 21 km and Ryan would have traveled 14 km.

The cost per highlighter is approximately $1.

To find the cost per highlighter, we can use a system of equations. Let's call the cost per highlighter "h" and the cost per pen "p". Then, we can set up two equations based on the information given;

80h + 20p = 120 (day 1)

120h + 40p = 200 (day 2)

We can simplify these equations by dividing both sides by 20;

4h + p = 6 (day 1)

6h + 2p = 10 (day 2)

Now we can use the first equation to solve for p in terms of h;

p = 6 - 4h

Substitute this expression for p into second equation;

6h + 2(6 - 4h) = 10

Simplify and solve for h;

6h + 12 - 8h = 10

-2h = -2

h = 1

Therefore, the cost per highlighter is $1.

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The value of theta in the given trigonometry function is 45⁰.

The value of theta in the given trigonometry function is calculated as follows;

9 sin²θ tanθ  -  9 sin²θ = 0

Solve the equation as follows;

9 sin²θ tanθ  =  9 sin²θ

divide both sides by  9 sin²θ;

(9 sin²θ tanθ)/9 sin²θ = 9 sin²θ /9 sin²θ

tanθ = 1

Now, solve for θ as follows;

θ = tan⁻¹ (1)

θ = 45⁰

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Answer: D

Step-by-step explanation: formal proofs deeply uses tables or a list of steps where informal proofs are proven in paragraphs.

water is being pumped into the tank at a rate of 23.7 cm³/min.we can take the derivative of the volume formula with respect to time and solve it

what is derivative  ?

In mathematics, a derivative is a measure of how much a function changes as its input value changes. More specifically, the derivative of a function is defined as the rate of change of the function at a particular point.

In the given question,

We can solve this problem using related rates, where we relate the rate of change of one variable to the rate of change of another variable.

Let's denote the height of the water in the tank by h, and the radius of the water surface by r. Then, we can use the formula for the volume of a cone to relate these variables:

V = (1/3)πr²h

We are given that the tank has height 9.0 m and diameter (and thus radius) 4.5 m at the top. We can use this information to find the relationship between h and r:

r = (1/2) × diameter = 2.25 m

h = 9.0 m - 2.0 m = 7.0 m

Now, we can take the derivative of the volume formula with respect to time:

dV/dt = (1/3)π(2r(dr/dt) + r²(dh/dt))

We want to find the rate at which water is being pumped into the tank, which is the rate of change of the volume with respect to time when the water level is rising at a rate of 21.0 cm/min. We are also given that water is leaking out of the tank at a rate of 11500.0 cm^3/min, so we can set these rates of change equal to each other:

dV/dt = (21.0 cm/min) × (1 m/100 cm)³ = 0.021 m³/min

11500.0 cm³/min = 11.5 × (1 m/100 cm)³ m³/min

Substituting these values and the values we found for r and h into the derivative formula, we get:

0.021 m³/min = (1/3)π(2(2.25 m)(dr/dt) + (2.25 m)²(11500.0 cm³/min)/(100 cm/m)³)

Simplifying and solving for dr/dt, we get:

dr/dt = [(0.021 m³/min) - (1/3)π(2(2.25 m)(11500.0 cm³/min)/(100 cm/m)³)] ÷ [(1/3)π(2(2.25 m))]

= 0.0237 m/min = 23.7 cm/min

Therefore, water is being pumped into the tank at a rate of 23.7 cm³/min.

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Area of the base of the pyramid is 16 square centimeters

Area of one lateral face of the pyramid  is  14 square centimeters

The lateral surface area of the pyramid is  56 square centimeters

The total surface area of the pyramid is 72  square centimeters

Area of the base of the pyramid = length x length

= 4 x 4

=16 square centimeters

Area of one lateral face of the pyramid = area of a triangle

=1/2×base×height

=1/2×4×7

=14 square centimeters

The lateral surface area of the pyramid =  4 x area of one lateral face

= 4 x 14

= 56square centimeters

The total surface area of the pyramid = lateral surface area of the pyramid +  area of the base

=56+16

=72 square centimeters

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Step-by-step explanation:

let the first letter of their name represent

their age

s=2j

h=j-5

s+j+h=35

replace s AND h with the values above

2j+j+j-5=35

4j=35+5

4j=40

j=10

replace j in the equation s=2j

s=2×10

s=20

replace j in the equation h=j-5

h=10-5

h=5

Answer

5

Solution

[5 (8^1/3 + 27^1/3)^3]^1/4

= [5 ((2^3)^1/3) + (3^3)^1/3)^3]^1/4

= [5((2+3)^3)1/4

= (5×5^3)^1/4

= (5^4)^1/4

= 5

Answer:  Approximately the year 2006

Roughly 8 years after 1998

================================================

Work Shown:

A = 373 represents a population of 373 thousand.

Plug in this value of A and solve for t. We'll need natural logs (LN) to isolate the variable.

[tex]A = 252e^{0.049t}\\\\373 = 252e^{0.049t}\\\\373/252 = e^{0.049t}\\\\1.4801587 \approx e^{0.049t}\\\\[/tex]

Apply natural logs to both sides.

[tex]\text{Ln}(1.4801587) \approx \text{Ln}\left(e^{0.049t}\right)\\\\\text{Ln}(1.4801587) \approx 0.049t*\text{Ln}\left(e\right)\\\\\text{Ln}(1.4801587) \approx 0.049t*1\\\\\text{Ln}(1.4801587) \approx 0.049t\\\\t \approx \text{Ln}(1.4801587)/0.049\\\\t \approx 8.0030472\\\\[/tex]

It takes about 8 years for the population to reach 373 thousand.

Since t = 0 starts at 1998, we get to the year 1998+8 = 2006.

The likelihood that a milkshake chosen at random will be a sloppy, thin, flavourless mess is 0.62.

The possibility or chance of an event occurring is measured by probability. It is a number between 0 and 1, where 0 denotes that the occurrence is impossible and 1 denotes that it is unavoidable. By dividing the number of favourable outcomes by the total number of potential outcomes, the probability of an occurrence is determined.

a. The milkshake will most likely turn out to be a watery, thin, flavourless mess (2, 4, 7).

The likelihood that a milkshake chosen at random would be a sloppy, thin, flavourless mess is equal to the final result of the milkshake will be a watery, thin, flavourless mess.

=2+4+7/21

=13/21

=0.62

b. No association can be removed from this table. We are unable to forecast the result of any event since the overall number of events is random.

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The measure of angle x of the circle is x = 5.7

Given data ,

Let the center of the circle be represented as x

Now , the value of x is given by

The measure of angle ∠VWU = 43°

And , the measure of arc VU = ( 7x + 46 )°

Now , from the arc theorem of circle , we get

mVU = 2 ∠VWU

On simplifying , we get

2  ( 43 ) = ( 7x + 46 )

7x + 46 = 86

Subtracting 46 on both sides , we get

7x = 40

Divide by 7 on both sides , we get

x = 5.71

Hence , the measure of x of circle is x = 5.7

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Hi! To determine if two lines are parallel given their equations, follow these steps:

1. Identify the slopes of the lines from their equations.

If the equations are in the slope-intercept form (y = mx + b), the slope is the coefficient of x (m). If the equations are in standard form (Ax + By = C), you can find the slope by rearranging the equation to the slope-intercept form.

2. Compare the slopes of both lines. If the slopes are equal, the lines are parallel.

In summary, given their math equations, you need to describe their slopes and check if they are equal to determine if the lines are parallel.

Hi to confirm that two lines are parallel lines given their equation you the gradient which is the value of x in the equation y=mx + c. Where m is the gradient

The two items that together cost 34% of her budget have a summed cost of $170

From the question, we have the following parameters that can be used in our computation:

She has a budget of $500.

To calculate which two items together cost 34% of her budget, we use

Amount = 34% * Budget

substitute the known values in the above equation, so, we have the following representation

Amount = 34% * 500

Evaluate

Amount = 170

Hence, the two items would have a summed cost of $170

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Answer:

2π(10^2) + 2π(10)(16)

= 2π(100) + 2π(160)

= 200π + 320π

= 520π ft^2

Answer: (4x - √3)(4x + √3).

Step-by-step explanation: Use factoring. The equation (when factored out) is: ax^2 + bx + c. Hence, the equation you wrote rewritten is: 16x^2 + 0x - 3. (√3)(√3) = 3 and (4x)(4x) = 16x^2. It is divisible.

So, the quadratic function f(x) = 16x^2 - 3 can be expressed as a product of linear factors as (4x - √3)(4x + √3).

Answer:w = 74.66 mph

d = 19.92 degrees

Step-by-step explanation:

Let's call the speed of the wind "w" and the direction angle of the wind "d". We can use vector addition to solve for these unknowns.

First, let's find the direction of the plane. The bearing of N * 65 deg * E can be represented as a vector with an initial point at the North pole and a terminal point 65 degrees east of due north. Since the plane is flying on this bearing, its direction vector is the same as the bearing vector. We can find the components of this vector as follows:

cos(65) = x / 1

x = cos(65)

y = sin(65)

So the direction vector of the plane is <cos(65), sin(65)>.

Next, let's find the ground speed vector. We can represent this vector as the sum of the plane's airspeed vector and the wind vector:

ground speed = airspeed + wind speed

We know that the magnitude of the ground speed vector is 390 mph, and we know the direction angle of the ground speed vector is 30 degrees. We can use this information to set up two equations:

|airspeed + wind speed| = 390

tan(d) = (wind speed)_y / (wind speed)_x

Since we know the direction vector of the plane, we can express the airspeed vector in terms of this vector as follows:

airspeed = 360 <cos(65), sin(65)>

Now we can substitute the airspeed vector and the wind speed vector into the equation for the ground speed vector, and use the fact that the magnitude of the ground speed vector is 390 to solve for the components of the wind speed vector:

|360 <cos(65), sin(65)> + <(wind speed)_x, (wind speed)_y>| = 390

Squaring both sides and using the fact that cos^2(x) + sin^2(x) = 1, we get:

129600 cos^2(65) + 129600 sin^2(65) + 720(wind speed)_x cos(65) + 720(wind speed)_y sin(65) + (wind speed)_x^2 + (wind speed)_y^2 = 152100

Simplifying, we get:

51840 + 720(wind speed)_x cos(65) + 720(wind speed)_y sin(65) + (wind speed)_x^2 + (wind speed)_y^2 = 152100

Substituting the equation for the direction angle of the wind and simplifying, we get:

51840 + 720w cos(65 - d) + w^2 = 152100 tan^2(d)

We now have two equations and two unknowns: w and d. We can solve for them using algebra or a numerical solver. Using a numerical solver, we find:

w = 74.66 mph

d = 19.92 degrees

Therefore, the speed of the wind is 74.66 mph, and its direction angle is 19.92 degrees.

Step-by-step explanation:

The equation of the line (by inspection) is y = 1/2 x

The number of feet of fencing that will be needed ​is illustrated below

The feet of fencing that will be needed ​is the perimeter of the fence

To find the perimeter of a composite figure, we need to add up the lengths of all the sides of the figure.

In this case, we add the side lengths of the rectangle to the side lengths of the right triangle and subtracting the common lengths between both shapes

Take for instance, we have

So, we have

Fencing = 2 * (4 + 5) + 3 + 4 + 5 - 2 * 4

Fencing = 22 units

So, the amount of fencing for the assumed dimensions is 22 units

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The valid row-operation which needs to be performed on the given matrix to get "1" in position of row1 and column1 are "R₁ → 2R₁ + R₂" and then "R₁ → R₁/2".

The "row-operation" is a type of matrix operation that involves swapping, scaling, or adding rows in a matrix. These operations are used to transform a matrix into a more reduced form.

The Augmented matrix on which we have to perform the "row-operation" is

⇒ [tex]\left[\begin{array}{ccc}7&-4&8\\-12&7&-13\end{array}\right][/tex],

To get "1" in first row and first column,

The first row-operation is R₁ → 2R₁ + R₂

We get,

⇒ [tex]\left[\begin{array}{ccc}7\times 2-12&-4\times 2+7&8\times 2-13\\-12&7&-13\end{array}\right][/tex],

⇒ [tex]\left[\begin{array}{ccc}14-12&-8+7&16-13\\-12&7&-13\end{array}\right][/tex]

⇒ [tex]\left[\begin{array}{ccc}2&-1&3\\-12&7&-13\end{array}\right][/tex],

The second , row-operation to get a "1", is R₁ → R₁/2,

We get,

⇒ [tex]\left[\begin{array}{ccc}2/2&-1/2&3/2\\-12&7&-13\end{array}\right][/tex],

⇒ [tex]\left[\begin{array}{ccc}1&-0.5&1.5\\-12&7&-13\end{array}\right][/tex].

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Answer: If you are trying to find the third derivative

It would be f'''(x)=159720(11x+3y)^3

Step-by-step explanation:

The third term of the given expression [tex](11x+3y)^6[/tex] is 135[tex]x^4y^2[/tex].

The binomial theorem states that [tex](a+b)^n[/tex] can be expanded as follows:

[tex](a+b)^n[/tex] = nC0*[tex]a^n[/tex] + nC1*[tex]a^{n-1[/tex]*b + nC2*[tex]a^{n-2[/tex]*[tex]b^2[/tex] + ... + nCn*[tex]b^n[/tex]

where nCk is the binomial coefficient, which is defined as n! / k!(n-k)!.

In this case, we have [tex](11x+3y)^6[/tex], so n = 6.

The third term of the expansion is given by nC2[tex]a^{n-2[/tex][tex]b^2[/tex], which is equal to 6C211[tex]x^4[/tex]3[tex]y^2[/tex].

6C2 is equal to 6! / 2!4! = 15, so the third term is equal to 1511[tex]x^4[/tex]3[tex]y^2[/tex] = 135[tex]x^4y^2[/tex].

Therefore, the third term of [tex](11x+3y)^6[/tex] is 135[tex]x^4y^2[/tex].

Here is a breakdown of the steps:

Find the binomial coefficient nC2.

Multiply nC2 by 11[tex]x^4[/tex] and 3[tex]y^2[/tex].

Simplify the expression.

The answer is 135[tex]x^4y^2[/tex].

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To explain why A + B is irrational, we need to show that it cannot be expressed as a ratio of two integers.

Suppose that A + B is rational, which means we can write it as the ratio of two integers p and q (where q is not zero):

A + B = p/q

Now, we can substitute the values of A and B and simplify:

√363 + √27 = p/q

We can then rearrange the terms to isolate one of the square roots:

√363 = p/q - √27

We can square both sides of this equation to eliminate the square roots:

363 = p^2/q^2 + 27 - 2(p/q)√27

Notice that the right-hand side of this equation has a term with a square root. This means that if we assume that A + B is rational, we arrive at a contradiction: we have shown that √363 (which is equal to A) is irrational, which means that p^2/q^2 + 27 must also be irrational. However, the left-hand side of the equation is rational. Therefore, our assumption that A + B is rational must be false, and we conclude that A + B is irrational.

To explain why AB is rational, we can use the fact that the product of two rational numbers is rational.

We can rewrite A and B as follows:

A = √(363) = √(121 x 3) = √(11^2 x 3) = 11√3

B = √(27) = √(9 x 3) = √(3^2 x 3) = 3√3

Therefore, AB = 11√3 x 3√3 = 33 x 3 = 99.

Since 99 is a rational number (which can be expressed as the ratio of the integers 99 and 1), we conclude that AB is rational.

Answer:

70...............................

Reason:

The left-most digit is the most significant when it comes to rounding. We bump the 6 up to 7 since 68 is closer to 70 (compared to 60). The other digits turn to 0 and become insignificant.

Do not write any of the following:

because that would mean we would have 2, 3, and 4 sig figs respectively. We simply write 70 without a decimal point.

The conditional statement is solved and you could ask for an enrollment verification notice from the school

Given data ,

Let the statement be informally proven that you are taking courses online

And , To informally prove that you are taking courses online, you can contact the professors of the courses via email and obtain informal permission from them first.

Then you can initiate the request using the new online information system of the school to obtain an enrollment verification notice

It is a legitimate document that the school has produced that may be utilized for a number of things, including debt deferments, employment, and insurance.

Hence , the conditional statement is solved

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