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3 years ago

2p + 2r = q solve for p

Mathematics

2 answers:

Gwar [14]3 years ago

4 0

Answer: P = 1/2q-r

Step-by-step explanation:

2p + 2r = q

2p = q - 2r

p = 1/2q - r

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Katena32 [7]3 years ago

3 0

Answer:

p=(q-2r)/2

Step-by-step explanation:

First, you must isolate the variable p, as you are solving for it. To do this, you substract 2r from both sides. Then, you are left with 2p=q-2r. To fully isolate p, you must divide both sides by 2. Lastly, you are left with p=(q-2r)/2.

Hope this helps :)

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3 years ago

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3 years ago

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3 years ago

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3 years ago

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

8 0

3 years ago

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