Ask question

Ask question

All categories

4 years ago

12

Point V lies between points U and W on line UW. The distance between U and V is 2x-4, and the distance between V and W is 4x+10.

If UW = 9x – 9, what is UW in units?

1 answer:

gogolik [260]4 years ago

8 0

Answer:

36UNITS

Step-by-step explanation:

If point V lies between U and W, then UV+VW = UW

given parameters

UV = 2x-4

VW = 4x+10

UW = 9x-9

Substituting the given functions in the formula first, we will have;

2x-4+4x+10 = 9x-9

6x+6 = 9x-9

Collect like terms

6x-9x = -9-6

-3x = -15

x = -15/-3

x = 5

Substitute x = 5 into UW.

Since UW = 9x-9

UW = 9(5)-9

UW = 45-9

UW = 36

You might be interested in

Find the missing ratio and explain how you found it

ankoles [38]

Answer:

5:4

Step-by-step explanation

If your shoes to flip-flops are 10:8 then the starting ratio will be 10:8 but 10 and 8 are both divisible to 2 so you can simplify 10:8 to 5:4

7 0

4 years ago

Read 2 more answers

I need help don't know what to do

givi [52]

Substitute 5 for g so...
5 times 11 =55

5 0

3 years ago

Read 2 more answers

Which of the following could not be a value of n? side lengths of 9 and 13

daser333 [38]

The correct answer here would be D. And that is because, in a triangle, the sum of 2 of the sides has to be greater than the third side for all sides. If we add 22 and 13 we get 35, and 35 is greater than 9, for sure. But if we add 13 and 9 we get 22 which is not greater than the third side length of 22; it would be collinear with the side length and it has to be greater than the side length.

8 0

4 years ago

DID CAROL KILL HER HUSBAND!!!!!!!!????<br> what do you think?

SSSSS [86.1K]

Answer:

maybe....

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

3 years ago