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Vesnalui [34]
4 years ago
11

I don't get how to solve for x with prathesis

Mathematics
1 answer:
Brums [2.3K]4 years ago
5 0
Just remember that something outside the parentheses applies to everything inside.

Examples:

3 times (x + 2) . . . . . means . . . . . 3 times 'x'  <u>plus</u>  3 times 2

a (b + c) . . . . . means . . . . . ab  <u>plus</u>  ac

4 (w + x + y + 2) . . . means . . . 4w + 4x + 4y + 8

When somebody gives you an equation that has parentheses in it,
the first thing you're always supposed to do is get rid of them. 
With the examples I just gave you, you know how to do that now.


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Please help me complete this !! 12points
Svet_ta [14]

Answer:

1.

x1=1

x2=3

y1=5

y2=1

midpoint= (x1+x2/2, y1+y2/2)=(1+3/2,5+1/2)=(2,3)

2.

x1=2

x2=4

y1=4

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(2+4/2,4+6/2)=(3,5)

3.

x1= -2

x2=4

y1=4

y2= -6

midpoint= (x1+x2/2, y1+y2/2)=(-2+4/2,4-6/2)=(1,-1)

4.

x1=0

x2=0

y1=3

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(0/2,3+5/2)=(0,4)

5.

x1= -2

x2=2

y1= -6

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(-2+2/2,-6+6/2)=(0,0)

6.

x1=1

x2=4

y1=4

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(1+4/2,4+5/2)=(5/2,9/2)

7.

x1=-3

x2=-4

y1=5

y2=5

midpoint= (x1+x2/2, y1+y2/2)=(-3-4/2,5+5/2)=(-7/2,5)

8.

x1=5

x2= -4

y1=2

y2=6

midpoint= (x1+x2/2, y1+y2/2)=(5-4/2,2+6/2)=(1/2,4)

3 0
3 years ago
What is the measure of an interior angle of a regular 9 sided polygon?
Talja [164]

Answer:

140

Step-by-step explanation:

3 0
3 years ago
Find the equation of the straight line which passes through (3,5) and makes y-inte<br>r-intercept.​
elixir [45]

Answer:

Please restate your question.

It needs the y-intercept to be decipherable.

Step-by-step explanation:

6 0
3 years ago
What is the answer!?!?!?
77julia77 [94]
The price for hardcover books is 1.50 and the price for paperback is .50
8 0
3 years ago
Limit   
STatiana [176]

Rationalize the numerator:

\dfrac{\sqrt{x+4}-2}x\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac x{x(\sqrt{x+4}+2)}=\dfrac1{\sqrt{x+4}+2}

This is continuous at x=0, so we can evaluate the limit directly by substitution:

\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}x=\lim_{x\to0}\frac1{\sqrt{x+4}+2}=\frac1{\sqrt4+2}=\frac14

5 0
3 years ago
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