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3 years ago

How do I solve this to find angle G?

Mathematics

1 answer:

VashaNatasha [74]3 years ago

5 0

Answer:

Step-by-step explanation:

The angle at E is 90 degrees.

All tangents of a circle always meet the radius at a 90 degree angle.

90 + 67 + G = 180o All triangles have 180 degrees. Combine left.

157 + G = 180o Subtract 157 from both sides.

157 - 157 + G + 180 - 157

G = 23

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What is the value of f(-5) ?

Salsk061 [2.6K]

Answer: 7

Step-by-step explanation:

f(x) = y

f(-5) = y

x = -5

find the value of y at x = -5

and the answer is 7

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2 years ago

Over a period of 12 weeks, Carlos ran an average of 21 miles per week. How many kilometers is this? Round your answer to the nea

velikii [3]

Answer:

405.80

Step-by-step explanation:

So, we know Carlos ran 21 miles per week, for 12 weeks....

That makes 21 mi/wk x 12 weeks = 252 miles.

Then we know each km is 0.621 mile.

So, we divide 252 miles by 0.621 mile/km to get= 405.80 km

Dividing miles by miles/km gives us the unit we want (km).

<u>There was a trap in this question,</u> because if instead of dividing the 252 miles by 0.621 you would have multiplied it, you would have gotten another answer listed... just not the right one. :-)

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3 years ago

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3 years ago

What must be the value of x?

tangare [24]

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3 years ago

Read 2 more answers

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

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3 years ago