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3 years ago

Does anyone get this?

Mathematics

1 answer:

stepladder [879]3 years ago

8 0

Answer: B

Step-by-step explanation:

In the systems of equation it already gives you the solution for x which is -2 so all you have to do is substitute -2 into the first equation and solve for y.

y= 2/3x + 3

x= -2

Substitute x for -2

y = $\frac{2}{3}* \frac{-2}1} + 3$

y= $\frac{-4}{3}$ + $\frac{3}{1}$

y = $\frac{5}{3}$

so now you have the solution like this $(-2,\frac{5}{3})$

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Johnny scored 8 out of 10 attempts, Ray scores 1 out of 2 attempts and David scored 12 out of 20 attempts

hjlf

Answer:

Johnny

Step-by-step explanation:

If we convert it so everyone has denominators of 20 we get:

(These are still equal to what their scores before were)

Johnny: 16/20

Ray: 10/20

David: 12/20

Johnny has made the most free throw shots, therefore, they have done the best.

Hope this helps!

-AxekickNebulite

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3 years ago

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R − ns × 12 for n = 2, r = 9, and s = 3.

valentina_108 [34]

Answer:

-63

Step-by-step explanation:

9 - (2×3) × 12

9 - (6 × 12)

9 - 72

= -63

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3 years ago

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Who can help me d e f thanks

12345 [234]

$y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}$

Product rule:

$y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'$

Chain and power rules:

$y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'$

Power rule:

$y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)$

Now simplify.

$y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}$

$y = \dfrac{3bx + ac}{\sqrt{ax}}$

Quotient rule:

$y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}$

$y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}$

Power rule:

$y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}$

Now simplify.

$y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}$

$y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}$

$y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}$

$y = \sin^2(ax+b)$

Chain rule:

$y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'$

$y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'$

$y' = 2a \sin(ax+b) \cos(ax+b)$

We can further simplify this to

$y' = a \sin(2(ax+b))$

using the double angle identity for sine.

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