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love history [14]

3 years ago

10

The triangle and the trapezoid have the same area. Base b2 is twice the length of base b1. What are the lenghts of the bases of

the trapezoid.

1 answer:

AfilCa [17]3 years ago

6 0

Kinda late, but do you have the area of the triangle?

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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

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

3 0

3 years ago

Find the values of x and y

zhuklara [117]

You can use a tangent:

$tangent=\dfrac{opposite}{adjacent}$

We have opposite = 17 and adjacent = x.

$\tan30^o=\dfrac{\sqrt3}{3}$

substitute:

$\dfrac{17}{x}=\dfrac{\sqrt3}{3}$ cross multiply

$x\sqrt3=(3)(17)$ multiply both sides by √3

$x(\sqrt3)(\sqrt3)=51\sqrt3$

$3x=51\sqrt3$ divide both sides by 3

$x=17\sqrt3$

Use the Pythagorean theorem:

$y^2=(17\sqrt3)^2+17^2\\\\y^2=289(\sqrt3)^2+289\\\\y^2=289\cdot3+289\\\\y^2=867+289\\\\y^2=1156\to y=\sqrt{1156}\\\\y=34$

-------------------------------------------------------------------------------------------------

Other method.

$30^o-60^o-90^o$ triangle.

The sides are in the ratio $1:2:\sqrt3\to17:y:x$

Therefore

$17:(2\cdot17):(17\sqrt3)\to17:34:17\sqrt3\to x=34,\ y=17\sqrt3$

4 0

3 years ago

What is 47.2 % of 300 ?

kupik [55]

For this type of question, you just need to multiply the percentage with the number
so, the calculation would be

47.2% x 300

47.2/100 x 300

47.2 x 3

= 141.6

6 0

3 years ago

Read 2 more answers

Please help me I don’t understand

DIA [1.3K]

Answer: A)

Explanation: 10-3=7

8 0

3 years ago

I’m practically begging someone please help me!!

Oliga [24]

My guys is is not just (5, -10)?

6 0

3 years ago