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2 years ago

What is the area of this trapezoid?

Mathematics

1 answer:

zavuch27 [327]2 years ago

7 0

Answer:

153

Step-by-step explanation:

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Fwee points UwU : DDDDD

SVETLANKA909090 [29]

Answer:

i got it yess

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

What is the answer?please help me

Maslowich

Answer:

y = 2x - 1 Graph B

y = 2x + 2 Graph A

y = 2x - 5 Graph C

Step-by-step explanation:

y = mx + c

c is where the line intersect the Y axis.

<h2>Substitute x for 0:</h2>

y = 2x - 1

2 * 0 - 1 = -1.

This is graph B because it intersects at 0,-1.

y = 2x + 2

2 * 0 + 2 = 2

This is graph A because it intersects at 0,2.

y = 2x - 5

2 * 0 - 5 = -5

This is graph C because it intersects at 0,-5.

6 0

2 years ago

Find the slope of line that passes through (5,6) and (7,9)

Debora [2.8K]

Answer: $\frac{3}{2}$

Step-by-step explanation:

m= $\frac{9-6}{7-5}$

= $\frac{3}{2}$

5 0

3 years ago

Find all excluded values for the expression.

ZanzabumX [31]

Answer:

-3

Step-by-step explanation:

For an expression to be undefined,. the denominator must be equal to 0

Therefore, in order to make that expression undefined, we must equate the denominator to 0

6 + 2v = 0

2v = 0 - 6

2v = -6

v = -6/2

v = -3

So in order to make the expression to be equal to 0, v must be -3

6 0

2 years ago

Find the taylor series for f(x) centered at the given value of a. [assume that f has a power series expansion. do not show that

Bond [772]

Taylor series is $f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }$

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

$f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }$

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

$f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }$

Following the pattern, we can see that for $f^{n}(x)$ ,

$f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} } \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}$

This applies for n ≥ 1, Expressing f(x) in summation, we have

$\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}$

Combining ln2 with the rest of series, we have

$f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }$

Taylor series is $f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }$

Find out more information about taylor series here

brainly.com/question/13057266

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3 0

1 year ago