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3 years ago

Help please solve for x.

Mathematics

1 answer:

Sedbober [7]3 years ago

4 0

Answer:

$x=\frac{60}{7}\ units$ or $x=8\frac{4}{7}\ units$

Step-by-step explanation:

<u><em>The correct question is</em></u>

The two rectangles show are similar. Find the value of x

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

In this problem

$\frac{20}{14}=\frac{x}{6}$

solve for x

$x=\frac{20}{14}(6)$

$x=\frac{120}{14}\ units$

simplify

$x=\frac{60}{7}\ units$

convert to mixed number

$x=\frac{60}{7}\ units=\frac{56}{7}+\frac{4}{7}=8\frac{4}{7}\ units$

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Write an equation for the line through (2,2), y-intercept 10

nasty-shy [4]

Y = -5x + 10
Your welcome for the help!

4 0

3 years ago

pleeeeeeeeeeeease hellllllllp meeeeeee the circle has a center of 0.its raduis is 3 cm,and the central angle A measures 100.what

Vsevolod [243]

Answer:

2.5 pi

Step-by-step explanation:

Comment

If you were trying to get the area of a whole circle, you would use

Area = pi r^2

You have to modify the formula to show that just part of the circle has an area that you are interested in.

The new formula is

Area = (theta/360) pi r^2

Givens

r = 30 cm

theta = 100

Solution

Area = (100 / 360) * pi * r^2 Substitute the givens into this formula

Area = (5 / 18) * pi * 3^2 Expand

Area = (5 / 18) * pi * 9 Cancel 9 into 18

Area = 5/2 * pi

Area = 2.5 * pi

7 0

2 years ago

The roof of the Terminal Tower is 709 feet above ground. A window washer dropped a bucket from the edge of the roof. How many se

Masja [62]

Answer:

6.64 seconds

Step-by-step explanation:

Given that:

Distance between roof of the terminal and ground = 709 feet

A bucket was dropped from the roof.

To find:

The time taken by the bucket to hit the ground?

Solution:

Distance traveled, $s$ = 709 feet

Initial velocity of the bucket, $u$ = 0 feet/sec (because initially it was at rest)

Acceleration of the bucket will be equal to acceleration due to gravity i.e.

$g$ or 32.1741 $ft/sec^{2}$

Let the time taken is $t$ seconds.

Formula for distance is given as:

$s = ut+\dfrac{1}{2}at^2$

Putting all the values:

$709 = 0 (t) + \dfrac{1}{2}(32.1741) t^2\\\Rightarrow 1418 = 32.1741 t^2\\\Rightarrow t^2 = 44.07\\\Rightarrow t \approx \bold{6.64\ sec}$

6 0

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

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

3 0

3 years ago

The area of a circle is 360.9. Find the radius of the circle. (Round to 2 decimal places.)

Ksju [112]

The radius of a circle is a=pi r^2.
so 360.9=3.14r^2, so that's around 114.94=r^2
r=10.72

4 0

3 years ago