All categories

2 years ago

Three sides of a quadrilateral are equal lengths. The length of the fourth side is 5 inches.

Mathematics

1 answer:

Korvikt [17]2 years ago

5 0

Answer:

Step-by-step explanation:

$x + x + x + 5 = 26$

$3x + 5 = 26$

You might be interested in

Helpppp plz will give brainliest

garik1379 [7]

11x = 44
X=4
44 + 7 = 51

5x = 20
31 + 20 =51

51 = 51

6 0

2 years ago

Solve by substitution x-y=-11 y+7=-2x

Bess [88]

x - y = - 11 .............( 1 )

y + 7 = - 2x ...............( 2 )

from equation ( 1 )

x - y = - 11

x = -11 + y ...........( 3)

putting x in equation ( 2 )

y + 7 = - 2 x

y + 7 = -2 ( -11 + y ) y + 7 = 22 - 2 y y + 2 y =22 - 7

3 y = 15 y = 15 / 3

putting value of y in equation 3

x = -11+ ( 13 /5 ) x = -33 /3 + 15 / 3 ( l.c.m)

x = -17 / 5

check

x = -11 + y

- 17 / 5 = -11 + 13 / 5

-17 /5 = -17 / 5

4 0

2 years ago

The area of trapezium is 1586cm² and the distance between sides is 26cm.If one of the parallel sides is 84cm,find the other

irinina [24]

Check the picture below.

we know the area is 1586, so

$\stackrel{\textit{area of a trapezoid}}{A=\cfrac{h(a+b)}{2}}\implies 1586=\cfrac{26(84+b)}{2}\implies 1586=13(84+b) \\\\\\ \cfrac{1586}{13}=84+b\implies 122=84+b\implies 38=b$

3 0

1 year 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

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

3 0

2 years ago

What is f(-2)?<br> -3 <br> -1<br> 1<br> 3

Dima020 [189]

Answer:

Step-by-step explanation:

ccccdcdc

7 0

2 years ago

Read 2 more answers