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3 years ago

What is the answer ?

Mathematics

1 answer:

Elan Coil [88]3 years ago

7 0

Answer:

A. AB≈ DE

is the answer to this question

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One number is 2 less than two thirds the other. their sum is 103. find the two numbers

anastassius [24]

One of the numbers is 40 and the other one is 63
confirmation:
63×(2/3)-2=40
and 40+63=103

6 0

4 years ago

Find the midpoint of the segment below and enter its coordinates as an ordered pair. If necessary, express the coordinates as fr

Vladimir [108]

The correct answer is (5/2, -7/2)

3 0

3 years ago

Dy/dx e^8x= sin (x+4y)

WINSTONCH [101]

Implicit differentiation:
differentiate both sides,
8e^(8x)=cos(x+4y)(1+4y')
Solve for y' to get
y'=[cos(x+4y)-8e^(8x)]/[4cos(x+4y)]

4 0

4 years ago

Brainliest if it is right

Natali [406]

Answer:

18.8

Step-by-step explanation:

multiply pi by the diameter (6 cm) and you get the circumference

hope that helps

8 0

3 years ago

Read 2 more answers

The following series are geometric series or a sum of two geometric series. Determine whether each series converges or not. For

Sergeeva-Olga [200]

Answer:

Required solution gives series (a) divergent, (b) convergent, (c) divergent.

Step-by-step explanation:

(a) Given,

$\sum_{n\to 0}^{\infty}\frac{2^n}{9^{2n}+1}$

To applying limit comparison test, let $a_n=\frac{2^n}{9^{2n}+1}$ and $b_n=\frac{9^{2n}}{2^n}$ . Then,

$\lim_{n\to\infty} \frac{a_n}{b_n}=\lim_{n\to\infty}(1+\frac{1}{9^{2n}})=1>0$

Because of the existance of limit and the series $\frac{9^{2n}}{2^n}$ is divergent since $\frac{9^{2n}}{2^n}=(\frac{9^2}{2})^n$ where $\frac{81}{2}>1$ , given series is divergent.

(b) Given,

$\sum_{n\to 1}^{\infty}(\frac{7^n}{7^n+4})$

Again to apply limit comparison test let $a_n=\frac{7^n}{7^n+4}$ and $b_n=\frac{1}{7^n}$ we get,

$\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{1}{7^n+4}=0$

Since $\lim_{n\to \infty} \frac{1}{7^n}=0$ is convergent, by comparison test, given series is convergent.

$\sum_{n\to 1}^{\infty}\frac{5^n+2^n}{6^n}= \sum_{n\to 1}^{\infty}(\frac{5}{6})^n+\sum_{n\to 1}^{\infty}(\frac{1}{3})^n$ . Now applying Cauchy Root test on last two series, we will get,

\lim_{n\to \infty}|(\frac{5}{6})^n|^{\frac{1}{n}}=\frac{5}{6}=L_1
\lim_{n\to \infty}|(\frac{1}{3})^n|^{\frac{1}{n}}=\frac{1}{3}=L_2

Therefore,

$\lim_{n\to \infty}\frac{5^n+2^n}{6^n}=L_1+L_2=1.16>1$

Hence by Cauchy root test given series is divergent.

5 0

4 years ago