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fredd [130]
3 years ago
6

Please help me solve this.​

Mathematics
1 answer:
erastovalidia [21]3 years ago
8 0

Answer:

B. Subtract 7

Step-by-step explanation:

You always want to isolate the <em>x</em> term to solve for <em>x</em> in any equation. Therefore, we need to move the 7 to the other side. Only way to do that is to subtract 7 on both sides.

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Can someone please help me on these questions!!!
34kurt

Answer:

#1 is 28

Step-by-step explanation:

Set the perimeters equal, as 6x - 2 = 8x - 12, solve for x = 5, and plug in.

3 0
3 years ago
Find the value of x that makes m|n.<br> m<br> (6x + 10)<br> (2x + 26)<br> The value of x is ?
olga nikolaevna [1]

Answer:

x is 9

Step-by-step explanation:

6x+10=2x+26

-2x          -2x

4x+10=26

   +10  +10

4x=36

---   ---

4    4

x=9

8 0
3 years ago
What is the fraction of 66.6%
puteri [66]

Answer:

333/500

Step-by-step explanation:

That's what it would be as a fraction.

6 0
4 years ago
Estimate the product 2 3/8 x 1/3
Aleonysh [2.5K]
The product of <span>2 3/8 x 1/3 will be found as follows:
We convert the first term into proper fraction
2 3/8=(2*8+3)/8=19/8
multiplying we get:
19/8</span>×1/3
=19/24

8 0
4 years ago
The coordinates of A, B, and C in the diagram are A(p,4), B(6,1), and C(9,q). Which equation correctly relates p and q?
siniylev [52]

Answer:

q+p=7

Step-by-step explanation:

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

In this problem line AB and line BC are perpendicular

so

m_A_B*m_B_C=-1

step 1

Find the slope AB

we have

A(p,4), B(6,1)

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

substitute the values

m=\frac{1-4}{6-p}

m_A_B=-\frac{3}{6-p}

step 2

Find the slope BC

we have

B(6,1), and C(9,q)

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

substitute the values

m=\frac{q-1}{9-6}

m_B_C=\frac{q-1}{3}

step 3

Find the equation that relates p and q

we know that

m_A_B*m_B_C=-1

we have

m_A_B=-\frac{3}{6-p}

m_B_C=\frac{q-1}{3}

substitute

(-\frac{3}{6-p})(\frac{q-1}{3})=-1

(\frac{q-1}{6-p})=1

q-1=6-p\\q+p=7

3 0
3 years ago
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