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3 years ago

Please answer both questions

Mathematics

2 answers:

tatuchka [14]3 years ago

6 0

A & C if I’m wrong I’m sorry

BaLLatris [955]3 years ago

5 0

Answer:

1)A

2) C

hope this helps you!

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Justice lived in Germany and Hollands for a total of 24 months in order to learn German. He learned and average of 120 words per

lesantik [10]

Answer:

Justice lived in Germany for 15 months and he lived in Holland for 9 months.

Step-by-step explanation:

1. You must make a system of equations:

$\left \{ {{x+y=24} \atop {120x+220y=3780}} \right.$

Where $x$ is the number on months he lived in Germany and $y$ is the number on months he lived in Holland.

2. You can solve this system of equations by using the Elimination Method. If you multiply the first equation by -120, you can cancel the variable $x$ when you add both equations, and then you can calculate $y$ . When you obtain the value of $y$ , you can substitute it into one of the original equations and solve for $x$ . Then:

$\left \{ {{-120x-120=-2880} \atop {120x+220y=3780}} \right.$

$100y=900\\y=9$

$x+y=24\\x+9=24\\x=15$

7 0

3 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

MariettaO [177]

The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3. Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then A(Y) = 2A(X) [T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3 But [T(X)]^2 = [A(X)]^3 Thus [T(Y)]^2 = 2^3[T(X)]^2 [T(Y)]^2 / [T(X)]^2 = 2^3 T(Y) / T(X) = 2^3/2 Therefore, the orbital period increased by a factor of 2^3/2 <span>
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6 0

3 years ago

Read 2 more answers

the length of a rectangular is 2 m less than twice the width.if the erea of the rectangular is 112m², what is the width

sergij07 [2.7K]

-7 or 8 is the answer

6 0

3 years ago

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.

7 0

2 years ago

Name

polet [3.4K]

Seth did. Seth scores 3/12 times and Zak scores 2/10 times. If you find a common denominator like 120, Zak scores 30/120 shots and Zak scores 24/120 shots. As a result Seth scores a greater fraction of shots.

8 0

3 years ago