Ask question

Ask question

All categories

goldfiish [28.3K]

3 years ago

8

1/4 | x + 2 | + 7 = 5

1 answer:

Katarina [22]3 years ago

7 0

Hope this is the correct answer! :)

You might be interested in

A line passes through (2, 9) and has a zero slope. Write an equation of the line satisfying the given conditions.

zubka84 [21]

Okay so for finding the equation of a line, which is a function, the equation is set to y, so you can eliminate A and B. When the slope of an equation is 0, we are saying that the y value of any point of the line is going to be the same, which is 9 from the given point (2,9).

So the answer is C. y=9

5 0

2 years ago

Help me please this is very important to me for rbainlist and explain

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

The probability of getting one even and one odd is the sum of probability of getting even and of getting odd and even. Since there are 3 even or odd on each die out of six numbers you have

P(EO)(3/6)(3/6)=1/4 and P(OE)=(3/6)(3/6)=1/4 so 1/4+1/4= 1/2

or you can find the probability of getting two odds or two evens and subtract that from 1

P(EE)=(3/6)(3/6)=1/4, P(OO)=(3/6)(3/6)=1/4

P(EO)=1-1/4-1/4=1/2

6 0

2 years ago

Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 5.00. What should be the length of B⃗ so that th

creativ13 [48]

Answer:

Length of B is 7.4833

Step-by-step explanation:

The vector sum of A and B vectors in 2D is

$C=A+B=(a_1+b_1,a_2+b_2)$

And its magnitude is:

$C=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2} =9$

Where

$a_1=Asinx$

$a_2=Acosx$

$b_1=Bsin(x+90)$

$b_2=Bcos(x+90)$

Using the properties of the sum of two angles in the sin and cosine:

$b_1=Bsin(x+90)=B(sinx*cos90+sin90*cosx)=Bcosx$

$b_2=Bcos(x+90)=B(cosx*cos90-sinx*sin90)=-Bsinx$

Sustituying in the magnitud of the sum

$C=\sqrt{(Asinx+Bcosx)^2+(Acosx-Bsinx)^2} =9$

$C=\sqrt{A^2sin^2x+2ABsinxcosx+B^2cos^2x+A^2cos^2x-2ABsinxcosx+B^2sin^2x} =9$

$C=\sqrt{A^2(sin^2x+cos^2x)+B^2(cos^2x+sin^2x)}$

$C=\sqrt{A^2+B^2} =9$

Solving for B

$A^2+B^2 =9^2$

$B^2 =9^2-A^2$

Sustituying the value of the magnitud of A

$B^2=81-5^2=81-25=56$

$B= 7. 4833$

8 0

3 years ago

What is 1/2=1/2c-2 plz show work

KatRina [158]

1/2 = 1/2c - 2

+2 + 2

2.5 = 1/2c

----- -----

1/2 1/2

5=c

3 0

3 years ago

Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

My name is Ann [436]

Answer:

Part 1) $BE=266\frac{2}{3}\ ft$

Part 2) $PE=233\frac{1}{3}\ ft$

Step-by-step explanation:

Part 1)

<u><em>Find the distance from B to E</em></u>

To solve this problem I assume that CGRP is a parallelogram

so

Triangles CGB and PEB are similar by AA Similarity Theorem

Remember that

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

so

$\frac{BG}{BE}=\frac{CB}{PB}$

substitute the values

$\frac{400}{BE}=\frac{300}{200}$

solve for BE

$BE=400(200)/300$

$BE=\frac{800}{3}\ ft$

Convert to mixed number

$BE=\frac{800}{3}\ ft=\frac{798}{3}+\frac{2}{3}=266\frac{2}{3}\ ft$

Part 2)

<u><em>Find the distance from P to E</em></u>

$\frac{CG}{PE}=\frac{CB}{PB}$

substitute the values

$\frac{350}{PE}=\frac{300}{200}$

solve for PE

$PE=350(200)/300$

$PE=\frac{700}{3}\ ft$

Convert to mixed number

$PE=\frac{700}{3}\ ft=\frac{699}{3}+\frac{1}{3}=233\frac{1}{3}\ ft$

5 0

3 years ago