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3 years ago

If WX is a midsegment of TUV find TU A 4 B 9 C 12 D 18

Mathematics

2 answers:

liubo4ka [24]3 years ago

7 0

Yes the answer is D 18

tigry1 [53]3 years ago

6 0

Answer:

is this a question meant to evaluate you, such as a test, exam, quiz, or assessment?

Step-by-step explanation:

the answer is d

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a rope is stretched from the top of a 7-foot tent pole to point on the ground 12 foot the base of the pole. how long is the rope

bulgar [2K]

Answer:

Step-by-step explanation:

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3 years ago

At a bike shop, Kirby is buying a new bike that costs

mart [117]

Answer:

tax: $18.9

total cost: $333.9

Step-by-step explanation:

$315 × 0.06 = 18.9

18.9 + 315 = 333.9

btw you can also do "315 × 1.06" to get the total cost without having to manually add the tax

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2 years ago

2 PART QUESTION!!

Llana [10]

Last answer for the first question and in not sure about the second one

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2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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2 years ago

Describe how the transformed function below is obtained from its parent function. y=x^2 4

Kazeer [188]

I believe its 4 we haven't really went over it

8 0

2 years ago

Read 2 more answers