Ask question

Ask question

All categories

3 years ago

15

A group of 8 girls went apple picking. The girls picked a total of 25 pounds of apples. They each get to take home the same amou

nt of apples. How many pounds of apples will each girl get?

1 answer:

valina [46]3 years ago

3 0

Answer:

Each girl will take home 3.125 poinds of apples

Step-by-step explanation:

25/8 = 3.125

You might be interested in

adrian wants to buy a set of 6 sets of wireless headphones he does not want to spend over $360.how much can he spend per set of

Alex787 [66]

$60 is the absolute maximum he can spend!

3 0

4 years ago

Read 2 more answers

If JKL - PNM. then M = L and the sides NP and<br> KJ are proportional.<br> True <br> Or<br> False???

nikklg [1K]

Answer:

True

Step-by-step explanation:

In similarity triangles, corresponding angles are congruent and corresponding sides are in proportion.

4 0

3 years ago

Read 2 more answers

1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

8 0

2 years ago

In art class mrsJocelyn gave everyone a 1 foot stick to measure and cut Vivian measured and cut her stick into 5 equal pieces Sc

garri49 [273]

Idk sorry i cant help

hope this helps XD

6 0

3 years ago

Answer please, thanks.

Rudik [331]

A) The perimeter is the length around the rectangle, so it would be 2 + 5 + 2 + 5 = 14m

The area is 2m * 5m = 10m^2

b/c) You have to draw a rectangle with a perimeter of 14m, and an area that ISN'T 10m^2. You could draw a 3m by 4m rectangle which would have a perimeter of 14m and an area of 12m^2.

4 0

4 years ago