Sally wants to fill 10, 8" tea glasses. How much tea does she need?

Can someone please answer this algebra question thanks!

Goryan [66]

Answer:

Yes it does answer A_____

6 0

3 years ago

For the composite function, identify an inside function and an outside function and write the derivative with respect to x of th

alexira [117]

Answer:

The inner function is $h(x)=4x^2 + 8$ and the outer function is $g(x)=3x^5$ .

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

Step-by-step explanation:

A composite function can be written as $g(h(x))$ , where $h$ and $g$ are basic functions.

For the function $f(x)=3(4x^2+8)^5$ .

The inner function is the part we evaluate first. Frequently, we can identify the correct expression because it will appear within a grouping symbol one or more times in our composed function.

Here, we have $4x^2+8$ inside parentheses. So $h(x)=4x^2 + 8$ is the inner function and the outer function is $g(x)=3x^5$ .

The chain rule says:

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)$

It tells us how to differentiate composite functions.

The function $f(x)=3(4x^2+8)^5$ is the composition, $g(h(x))$ , of

outside function: $g(x)=3x^5$

inside function: $h(x)=4x^2 + 8$

The derivative of this is computed as

$\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=3\frac{d}{dx}\left(\left(4x^2+8\right)^5\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=u^5,\:\:u=\left(4x^2+8\right)\\\\3\frac{d}{du}\left(u^5\right)\frac{d}{dx}\left(4x^2+8\right)\\\\3\cdot \:5\left(4x^2+8\right)^4\cdot \:8x\\\\120x\left(4x^2+8\right)^4$

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

3 0

4 years ago

jill traveled to the ferry office and back. the trip there took 5 hours and the trip back took 4 hours. she averaged 65 mph on t

antoniya [11.8K]

We know from physics class that the formula for distance of a linear motion is given as:

d = v t

Where,

d = distance travelled

v = average velocity

t = time it took to reach the destination

Since the distance going to the office and back is just similar, therefore we can simply equate the two:

v1 t1 = v2 t2

Where 1 signifies going to the office and 2 signifies going back from the office. Therefore this yields to:

v1 * 5 hours = 65 mph * 4 hours

v1 = 52 mph

<span>Answer: The average speed going to the office is 52 mph.</span>

4 0

3 years ago

Can someone help me with both of the or may be one

azamat

Question 6

Given:

QR = RS

QR = x + 6

RS = 4x

To find:

Length of line segment QS

Steps:

We know QR = RS, so substituting we get,

x + 6 = 4x

6 = 4x - x

6 = 3x

6/3 = x

2 = x

x = 2

Now,

QS = QR + RS

QS = x + 6 + 4x

QS = 2 + 6 + 4(2)

QS = 2 + 6 + 8

QS = 8 + 8

QS = 16 units

Therefore, the length of QS is 16 units

Question 7

Given:

QR = RS

QR = 2x - 2

RS = 2x

To find:

Length of line segment QS

Steps:

We know that QR = RS, so substituting the values we get,

QR = RS

3x - 2 = 2x

3x - 2 - 2x = 0

3x - 2x = 2

x =2

Now,

QS = QR + RS

QS = 3x - 2 + 2x

QS = 3(2) - 2 + 2(2)

QS = 6 - 2 + 2(2)

QS = 6 - 2 + 4

QS = 4 + 4

QS = 8 units

Therefore, the length of QS is 8 units

Happy to help :)

If u need any help feel free to ask

7 0

3 years ago

Read 2 more answers

Is andies solution correct

Nadusha1986 [10]

The solution is correct.

3 0

3 years ago