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3 years ago

10

Six angles share a vertex. Each of the angles has the same measure. The sum of the measures of the angles is 330 degrees. What i

s the measure of one angle?

1 answer:

JulsSmile [24]3 years ago

8 0

The measure of one angle is 55 because 330 divided by six is 55

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Combine like terms to simplify 3x + 3 + 3x + 8=

Over [174]

Answer:

6x + 11

Step-by-step explanation:

3x + 3 + 3x + 8

Combine like terms (terms with the same amount of variables).

3x + 3x + 3 + 8 =

(3x + 3x) + (3 + 8) =

6x + 11

6x + 11 is your answer.

~

3 0

3 years ago

Find the measure of each missing angle

podryga [215]

Answer:

ok sooo....

Step-by-step explanation:

angle 1=180-118=62

angle2=62+95=157 180-157=23

angle3=180-23=157

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2 years ago

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Lynna [10]

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3 years ago

Linda is sending out cards.If envelopes come in boxes of 25 and stamps come in packs of 10,what is the least number of stamps an

chubhunter [2.5K]

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3 0

3 years ago

If x = a cosθ and y = b sinθ , find second derivative

Olin [163]

I'm guessing the second derivative is for <em>y</em> with respect to <em>x</em>, i.e.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}$

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$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

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$x=a\cos\theta\implies\dfrac{\mathrm dx}{\mathrm d\theta}=-a\sin\theta$

and so

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{b\cos\theta}{-a\sin\theta}=-\dfrac ba\cot\theta$

Now compute the second derivative. Notice that $\frac{\mathrm dy}{\mathrm dx}$ is a function of $\theta$ ; so denote it by $f(\theta)$ . Then

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm dx}$

By the chain rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm df}{\mathrm d\theta}\dfrac{\mathrm d\theta}{\mathrm dx}=\dfrac{\frac{\mathrm df}{\mathrm d\theta}}{\frac{\mathrm dx}{\mathrm d\theta}}$

We have

$f=-\dfrac ba\cot\theta\implies\dfrac{\mathrm df}{\mathrm d\theta}=\dfrac ba\csc^2\theta$

and so the second derivative is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\frac ba\csc^2\theta}{-a\sin\theta}=-\dfrac b{a^2}\csc^3\theta$

4 0

3 years ago