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2 years ago

5

The diameter of a circle is 3 times the side length of a square . If the square has an area if 144, what is the length of the ra

dius in the circle?

1 answer:

stellarik [79]2 years ago

5 0

If the square's area is 144, then the two sides are both 12.
The diameter is 3 times the side length (12): 3*12=36
The radius is half of the diameter: 36/2=18

The radius is 18

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Which expressions below are equivalent to

Yuri [45]

Answer:

<h2>A, B, D, E</h2>

Step-by-step explanation:

2(2x + 1) = (2)(2x) + (2)(1) = 4x + 2 → A

<em>used distributive property</em>

2(2x + 1) = 2(1 + 2x) → B

<em>used commutative property</em>

2(2x + 1) = (2x + 1) + (2x + 1) = 2x + 1 + 2x + 1 → D

<em>used 2a = a + a</em>

2(2x + 1) = 4x + 2 = x + x + x + x + 1 + 1 → E

<em>used 4x = x + x + x + x and 2 = 1 + 1</em>

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2 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
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and so on.

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3 years ago

We gave a national verbal proficiency test that has a mean score of 500 with a standard deviation of 75. What raw score would co

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Answer:

519

Step-by-step explanation:

Let's assume that the scores for the national verbal proficiency test follow a normal distribution.

Mean score (μ) = 500

Standard deviation (σ) = 75

According to a Z-score table, the score for the 60th percentile is roughly z = 0.253

For any score "x", the z-score is given by:

$Z=\frac{X-\mu}{\sigma}$

For z = 2.53:

$0.253=\frac{X-500}{75}\\X=519$

The raw score to the 60th percentile is 519.

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Answer:

Step-by-step explanation:

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3 years ago