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3 years ago

Can you help me please...

Mathematics

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

-9

Step-by-step explanation:

hope that helps dhdh

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Paolo paid $ 28 $ 28 for a hat that was originally priced at $ 35 $ 35 . By what percent was the hat discounted?

devlian [24]

The original price of the hat is $35. Paolo paid $28 for the hat.

So discount for the hat is $(35-28) = $7

$7 is discounted from $35. We have to find the percentage of the discount.

To find the percentage we have to divide the discount amount by the original price and then have to multiply it by 100.

So the discount percentage =

((7/35)×100) %

We can simplify 7/35 by dividing 7 and 35 both by 7. So we will get 1/5 after simplifying.

((1/5) ×100) %

(100/5)% = 20%

So the hat was discounted by 20%.

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3 years ago

Elsa scores eight points in six minutes if she could send years at this rate how many points will she score 9 minutes

Minchanka [31]

Answer:

12 points

Step-by-step explanation:

9 is 1.5 times 6, so 8 times 1.5 gives you the answer of 12 points.

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3 years ago

What is the x and y intercept for 4x-5y=-20

yuradex [85]

X intercept is when y = 0 so 4x = -20, x=-5
y intercept when x = 0 so -5y = -20, y=4

7 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,

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