What is the volume of the composite figure?

Answer:

A,D,C

Step-by-step explanation:

We can solve this question using the formula for compound interest:

F = ?

P = $2500

i = 1.75

n = 4

Then, we have:

The value of her account after 4 years is about $2679.65.

Answer:

Not a function

Step-by-step explanation:

This is not a function. It will fail the vertical line test.  For any value of x, there are two different y values.  This tells us that it is not a function

Answer:

spped of the plane in calm air=121 km/h

speed of the wind= 4km/h

Step-by-step explanation:

Let say V the speed of the plane in calm air

and v the speed of the wind

Flying with a tailwind, a flight crew flew 500 km in 4 hours ==> 500= (V+v)*4

Flying against the tailwind, the crew flew 468 km in 4 hours ==> 468 = (V-v)*4

We divide the 2 equations by 4 and then add the 2 results equations:

(500+468)/4=2V ==> V=121 (km/h)

We replace that value in the first equation:

V+v=500/4=125

v=125-121=4 (km/h)

The particle travels approximately 45.63 units along the path. The displacement is the straight-line distance between the initial and final positions of the particle is 2781.

To find the distance the particle travels along the path, we can integrate the speed over the interval of time. The speed of the particle is given by the magnitude of its velocity vector.

The velocity vector is the derivative of the position function r(t):

\(v(t) = (d/dt)(t^3/3, t^2, 2t)\)

\(= (t^2, 2t, 2)\)

The speed of the particle at any given time t is:

|v(t)| = √((t^2)^2 + (2t)^2 + 2^2)

= √(t^4 + 4t^2 + 4)

= √((t^2 + 2)^2)

To find the distance traveled along the path, we integrate the speed function over the given interval of time. The particle travels from t = -1/3 to t = 9.

distance = ∫[from -1/3 to 9] |v(t)| dt

= ∫[from -1/3 to 9] |t^2 + 2| dt

= ∫[from -1/3 to 0] -(t^2 + 2) dt + ∫[from 0 to 9] (t^2 + 2) dt

= [-1/3 * t^3 - 2t] (from -1/3 to 0) + [1/3 * t^3 + 2t] (from 0 to 9)

Evaluating the definite integrals:

distance = [-1/3 * 0^3 - 2 * 0 - (-1/3 * (-1/3)^3 - 2 * (-1/3))] + [1/3 * 9^3 + 2 * 9 - (1/3 * 0^3 + 2 * 0)]

= [0 - (1/3 * (-1/27) + 2/3)] + [1/3 * 729 + 18]

= [1/27 + 2/3] + [729/3 + 18]

= 1/27 + 2/3 + 729/3 + 18

= 1/27 + 18/27 + 729/3 + 18

= (1 + 18 + 729)/27 + 18

= 748/27 + 18

= 27.63 + 18

= 45.63 units (approximately)

Therefore, the particle travels approximately 45.63 units along the path.

To find the average speed, we divide the distance traveled by the time taken. The time taken is 9 - (-1/3) = 9 1/3 = 28/3.

average speed = distance / time

= 45.63 / (28/3)

= 45.63 * (3/28)

= 4.9179 units per unit time (approximately)

The displacement is the straight-line distance between the initial and final positions of the particle.

displacement = |r(9) - r(-1/3)|

= |(9^3/3, 9^2, 2 * 9) - ((-1/3)^3/3, (-1/3)^2, 2 * (-1/3))|

= |(27, 81, 18) - (-1/27, 1/9, -2/3)|

= |(27 + 1/27, 81

= 2781.

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The requried solution of the given derivative has been given below.

If F(t) is an antiderivative of t⁵, then we have:

F'(t) = d/dt(F(t)) = t⁵

According to the Second Fundamental Theorem of Calculus, we have:

\(\int_1^x t^5 dt = F(x) - F(1)\)

Using this formula, we can find \(d/dx(\int_1^x t^5 dt)\) as follows:

\(d/dx(\int_1^x t^5 dt) = d/dx(F(x) - F(1)) = F'(x) - 0 = t^5\)

Therefore, \(d/dx(\int_1^x t^5 dt) = t^5.\)

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Set up the equation to determine the arc length. Arc length is equal to 2 pi times the arc's radius times (arc's central angle/360), where arc's central angle is expressed in degrees.

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\(x7 - 2 \: or \: x \leqslant - 3\)

Answer:

6.8333333333 or 41/6

Step-by-step explanation:

The equivalent decimal and mixed number for given letter C is,

⇒ C = 8.2 ; 8 2/10

We have to given that,

A number line is shown in image.

Since,

A number lines are the horizontal straight lines in which the integers are placed in equal intervals.

And, All the numbers in a sequence can be represented in a number line. This line extends indefinitely at both ends.

Now, By given number line,

Point C is two point left from point 8.

Hence, The point C is denoted as,

⇒ C = 8.2

⇒ C = 82/10

⇒ C = 8 2/10

Therefore, the equivalent decimal and mixed number for given letter C is,

⇒ C = 8.2 ; 8 2/10

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Answer: Area 6 Perimeter 12

Step-by-step explanation:

First, you have to find the perimeter. 4+5+3=12. Then, you need to find the area, which is base*height/2. The base is 3 and the height is 4. 4*3=12. Then you need to do 12/2=6. The area is 6

Answer:

The answer for P=12cm,A=6cm²

Step-by-step explanation:

Perimeter of triangle=a+b+c

P=5+3+4

P=12cm

Area of triangle =1/2bh

A=1/2×3×4

A=2×3

A=6cm²

Answer:

45 is answer I guess cuz my teacher taught me just like that

let's take a peek at the picture above, hmmm let's notice the vertex is at (-1 , 2), now let's get a point besides the vertex hmmm let's see it passes through (-2 , -1).

So we can reword that as what's the equation of a quadratic whose vertex is at (-1 , 2) and it passes through (-2 , -1)?

\(~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{a~is~negative}{op ens~\cap}\qquad \stackrel{a~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\)

\(\begin{cases} h=-1\\ k=2\\ \end{cases}\implies y=a(~~x-(-1)~~)^2 + 2\hspace{4em}\textit{we also know that} \begin{cases} x=-2\\ y=-1 \end{cases} \\\\\\ -1=a( ~~-2-(-1) ~~ )^2 + 2\implies -3=a(-2+1)^2\implies -3=a \\\\\\ ~\hfill~ {\Large \begin{array}{llll} y=-3(x+1)^2 + 2 \end{array}} ~\hfill~\)

Answer:

y = -3(x + 1)^2 + 2

Step-by-step explanation:

y = a(x - h)^2 + k is the vertex form of a quadratic, where

Finding (h, k):

We see from the graph that the vertex is a maximum and its coordinates are (-1, 2).  Thus h is -1 and k is 2.  Since h becomes negative, it will be 1 in the parentheses: (x - (-1) = (x + 1).

Finding a:

In order to find a, we will need to plug in a point on the parabola for (x, y) and (-1, 2) for h and k.  We see that (0, -1) lies on the parabola so we can use this point for (x, y).

-1 = a(0 - (-1))^2 + 2

-1 = a(0 + 1)^2 + 2

-3 = a(1)^2

-3 = a

Thus, a = -3.

Thus, the exact equation in vertex form of the parabola is:

y = -3(x + 1)^2 + 2

I attached a picture from Desmos Graphing Calculator that shows how the equation I provided works and contains the two points you marked on the parabola, including (-1, 2) aka the maximum, and (0, -1) aka the y-intercept.

A rotating sprinkler that sprays water at a radius of 5 feet is used to water a lawn. The area of the lawn that is being watered is: 78.55 feet². This can be solved by using area of circle.

The area of a circle is π multiplied by the square of the radius. The area of a circle (A) when the radius 'r' is given is πr². π is approximately  3.142

here, we have,

area = π x r²

radius = 5

thus, 5² = 25

= 25 x 3.142

= 78.55 feet²

Hence, The area of the lawn that is being watered is: 78.55 feet².

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Answer:

Congruent by AAA

Step-by-step explanation:

Given

\(\triangle MNP\)

\(\triangle QST\)

Required

The congruence rule

From the attached triangles, we have:

\(\angle M = \angle Q\) --- A

\(\angle N = \angle S\) --- A

\(\angle P = \angle T\) --- A

Hence, the congruence rule is AAA

Answer:

\(y^{-9}\neq \dfrac{1}{y^{-9}}\)

Step-by-step explanation:

Tyler simplified the expression: \(x^{-3}y^{-9}\)  as shown below:

\(x^{-3}y^{-9}=\dfrac{1}{x^3}X \dfrac{1}{y^{-9}}=\dfrac{1}{x^3y^{-9}}\)

We notice that in Tyler's work:

\(y^{-9}= \dfrac{1}{y^{-9}}\)

Whereas, the correct form would have been:

\(y^{-9}= \dfrac{1}{y^{9}}\)

Making our solution:

\(x^{-3}y^{-9}=\dfrac{1}{x^3}X \dfrac{1}{y^{9}}=\dfrac{1}{x^3y^{9}}\)

Answer:

1/53 I believe because it says I believe sorry if I am wrong hope you get it right

Answer:

u'=2(2t³-21t)³(6t-21)

Step-by-step explanation:

For a. they want you take the derivative of that function but in terms of t, which means you have to plug in that x=... part at some point.

You can either plug it in first and then take a derivative or take the derivative first then plug it in. But remember if you take a derivative first then u'=2x³(x') due to the chain rule; so you'd need to plug in the derivative of x anyways. It's really just a matter of preference at that point.

Doing it either way gives us:

u'=2(2t³-21t)³(6t-21)

For b. you just need to plug in 2 and simplify to get your answer there.

Answer:

The group of students on the top right those students have a mean of 2.6

Step-by-step explanation:

The mean is adding up all the different numbers of pets for each kid and dividing it by the number of kids so 13 the number of pets and 5 the number of different students

The true statement is if k = - 1 there are solutions, but if k = - 3 there are no solutions

Given: |x+3|-2=k

Below are rules for solving absolute value equations:

1. If p is positive and |y| = p, then

x = p   OR    y = -p

(two equations are set up)

2. If p is negative and |y| = p, then

No solution

3. If p is zero  and |y| = p, then

One solution

Using the rules:

|x+3|-2=k

when k = -1

|x+3|-2= -1    

|x+3| = 1

Rule 1 is applicable here. Thus, there are solutions  

when k = -3

|x+3|-2 = -3

|x+3| = -1

Rule 3 is applicable here. Thus, no solution

Therefore, if k = - 1 there are solutions, but if k = - 3 there are no solutions.     The 2nd option is the true statement

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Answer:

8.35

Step-by-step explanation:

To get absolute value, just make it the opposite

Answer: |-8.35| = 8.35

Step-by-step explanation: |-8.35| = 8.35

Answer:

10 i think thats not really a questuon tho

Answer:

\(\boxed{m = \frac{5}{2}}\)

.

Step-by-step explanation:

Use the form below

\(\boxed{\boxed{m = \frac{y_2 - y_1}{x_2 - x_1} }}\)

Where

.

So, the slope is

\((-5,~ 3) \to x_1 = -5~and~y_1 = 3\)

\((-3,~ 8) \to x_2 = -3~and~y_2 = 8\)

.

\(m = \frac{8-3}{-3-(-5)}\)

\(m = \frac{5}{-3+5}\)

\(m = \frac{5}{2}\)

.

Happy to help:)

Answer:

jwjs sjsj sjaj sjja sjja sjja sjja sjjaa

Step-by-step explanation:

jwjjw wjwj wjwjw wjwjw jwjw wjwbwh wbwb iwbwb wbjwbwv whvwwvwhwvwbvw w w wvw w whww w

we have a maximum at t = 0, where the maximum is y = 30.

We have a minimum at t = -1 and t = 1, where the minimum is y = 20.

These are given by the zeros of the first derivation.

In this case, the function is:

w(t) = 10t^4 - 20t^2 + 30.

The first derivation is:

w'(t) = 4*10t^3 - 2*20t

w'(t) = 40t^3 - 40t

The zeros are:

0 = 40t^3 - 40t

We can rewrite this as:

0 = t*(40t^2 - 40)

So one zero is at t = 0, the other two are given by:

0 = 40t^2 - 40

40/40 = t^2

±√1 = ±1 = t

So we have 3 roots:

t = -1, 0, 1

We can just evaluate the function in these 3 values to see which ones are maximums and minimums.

w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20

w(0) = 10*0^4 - 20*0^2 + 30    = 30

w(1) =  10*(1)^4 - 20*(1)^2 + 30 =  20

So we have a maximum at x = 0, where the maximum is y = 30.

We have a minimum at x = -1 and x = 1, where the minimum is y = 20.

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Answer:

7, 11, 15, 19, 23, 27, 31... and keep going until you have an answer.

Step-by-step explanation:

Just keep adding 4 until you have all the 48 terms you need.

Answer:

How much pizzas were sold cuz there is no way to solve it without knowing how much she sold

Step-by-step explanation:

Answer:

Step-by-step explanation:

Letter B is the correct answer

Initial population is 200, tripled every hour

200 x 3 = 600

First hour 200

Second hour = 600 + 200 = 800

Third hour 800x3 = 2400 + 800 = 3200

and so on.