All categories

3 years ago

12 dollars for 3 paper back books and 28 dollars for 7 paper back books are they equivalent please explain

1 answer:

3 0

If you have 12 dollars and you buy 3 paper back books, they each cost the same amount. So you would divide 12 dollars among 3 books to get the same answer for all of them. 12 divided by 3 equals the answer.

Then if you have 28 dollars and you buy 7 paper back books, they all cost the same amount. So that would be 28 divided by 7, equals the same answer.

You might be interested in

Is 5 5/6 + 1 3/4=5 1/12 reasonable

ira [324]

Answer:

No, it is unreasonable.

Step-by-step explanation:

5/5/6+1/3/4=35/6+7/4

=70/12+21/12

=91/12

=7 7/12

4 0

3 years ago

What's y in: 21y=(-205) +75+47y

Alenkinab [10]

The Answer Is A Fraction y= 151/47

6 0

3 years ago

Read 2 more answers

What is the value of x, the side length of the smallest square (please show work) Thanks!

Ratling [72]

Answer:

8 cm

Step-by-step explanation:

Ok lets on the biggest square we have 17 cm

17cm is the hypotenuse for the right triangle

The second largest's area is 225 cm^2

Sq root of 225 cm^2, is 15

To find x, lets plug the info we have to the pythagorean theorom

$a^2 + b^2 = c^2\\a^2 + 15^2 = 17^2\\a^2 + 225 = 289\\\\289 - 225 = a^2\\64=a^2\\a = 8 cm$

The side length of the smallest Square is 8 cm

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

7 0

3 years ago

The angle shown is in standard position. Select the possible measures of the angle. –450° –360° –270° –90° 90° 270° 360° 450° On

erma4kov [3.2K]

Answer:

-450

-90

270

Step-by-step explanation: Just did it on edgen-uity

4 0

3 years ago

Read 2 more answers

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago