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3 years ago

5

Enter answer in the space below

2 answers:

almond37 [142]3 years ago

5 0

Put the rest of the info too so ppl could solve it lolll

ira [324]3 years ago

3 0

We need more info to solve it

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I know the answers but i need the work. plz help

inysia [295]

For number 3.

You add 8 to both sides as you need to separate the variable

4x-8=32
+8 +8

4x=32

now divide 32 by 4

x=8

for number 4

you use pemdas to multiply the parenthisis

2(x-5)=-20

(2•x)-(2•5)

2x-10=-20

then add 10 to both sides

2x-10=-20
+10 +10

2x=10

divide 10 by two

x=5

5 0

3 years ago

Solve for X: 2-14x+52=7

meriva

Answer:

Step-by-step explanation:

x=

47

14

3 0

3 years ago

Read 2 more answers

5. Determine whether the

BARSIC [14]

Step-by-step explanation:

5. Determine whether the

x

y

11

-2

-1

2.

0

-1

1

2

2

11

Is it a linear quadratic or exponential

6 0

3 years ago

Nadia is investigating rotations about the center of regular polygons that carry the regular polygon onto itself. She claims tha

GalinKa [24]

Answer:

$\theta_1 \ n\ \theta_2 = 120, 240$

Step-by-step explanation:

The question is incomplete, as the angles of rotation are not stated.

However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself

We have:

$Nonagon = 9\ sides$

$Hexagon = 6\ sides$

Divide 360 degrees by the number of sides in each angle, then find the multiples.

<u>Nonagon</u>

$\theta = \frac{360}{9} =40$

List the multiples of 40

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

<u>Hexagon</u>

$\theta = \frac{360}{6} =60$

List the multiples of 60

$\theta_2 = 60, 120, 180, 240, 300$

List out the common angles

$\theta_1 = 40, 80, 120, 160, 200, 240, 280, 320$

$\theta_2 = 60, 120, 180, 240, 300$

$\theta_1 \ n\ \theta_2 = 120, 240$

This means that, only a rotation of $120, 240$ will lift both shapes onto themselves, when applied to both shapes.

The other angles will only work on one of the shapes, but not both at the same time.

7 0

3 years ago

Y>1 and y>x on a graph

zlopas [31]

So,

$y>1\wedge y>x$

1. Graph each inequality separately.

2. Choose a test point to determine which side of the line needs to be shaded.

3. The solution to the system will be the area where the shadings from each inequality overlap one another (purple area)

As for the system of inequalities we say it's unbounded.

3 0

3 years ago