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vitfil [10]
3 years ago
12

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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R − ns × 12<br><br><br> for n = 2, r = 9, and s = 3.
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Answer:

-63

Step-by-step explanation:

9 - (2×3) × 12

9 - (6 × 12)

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3 years ago
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Who can help me d e f thanks​
12345 [234]

d)

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}

e)

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}}

f)

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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