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andrey2020 [161]
3 years ago
5

Use the segment addition postulate to solve for BC when AB = 37 and AC

Mathematics
1 answer:
Ivan3 years ago
4 0

Answer:

BC = 56

Step-by-step explanation:

Segment Addition Postulate: AB + BC = AC

  1. 37 + BC = 93
  2. Subtract 37 from both sides
  3. BC = 56

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Maggie took one of her friends out to lunch. The lunches cost $20 and she paid 5% sales tax. If Maggie left a 10% tip on the $20
RUDIKE [14]

Answer:

46$

Step-by-step explanation:

20+20=40, 5% of 40=2, 42, +4, 46. (the 4 comes from the 10%)

7 0
3 years ago
I need help please?!!!!!!
Sindrei [870]

Answer:

-2x-10 ≥ 22

Collect like terms

-2x ≥ 22+10

-2x ≥ 32

divide both sides by -2

x ≥ -16.

5 0
3 years ago
What is (x-4)(2x-3)=0?
amm1812
Hi,
The answer is 3/2 or 4
How I got my answer:
Just think that (2x+3) OR (x-4) has to be 0.
<span> Than you can find it out with a simple +/- calculation. </span>
<span> 2x+3=0 x-4=0 
x=-3/2 or 4</span>Hope this helps you.
7 0
3 years ago
Pls help me in these two little exercises about the slope.
den301095 [7]

Answer:

3)  y=\dfrac35x+\dfrac25

4) a)  y=-2x+7

  b)  y=\dfrac12x+\dfrac92

Step-by-step explanation:

<u>Exercise 3</u>

-3x + 5y = 2

\implies 5y = 3x + 2

\implies y=\dfrac35x+\dfrac25

<u>Exercise 4</u>

a) If L2 is parallel to L1, it has the same slope (gradient) ⇒ m = -2

If L2 passes through point (3, 1):

y-y_1=m(x-x_1)

\implies y-1=-2(x-3)

\implies y=-2x+7

So L2 = L1

b) If L3 is perpendicular to L1, then the slope of L3 is the negative reciprocals of the slope of L1  ⇒  m = \dfrac12

If L3 passes through point (-5, 2):

y-y_1=m(x-x_1)

\implies y-2=\dfrac12(x+5)

\implies y=\dfrac12x+\dfrac92

4 0
2 years ago
Read 2 more answers
How would you solve this by elimination?
jekas [21]
To solve this equation by elimination, what you would do is multiply one of the equations by -1, or distribute -1 to each term in the equation, any of the 2 equations. Then align the equations and add them together.

-(X + 3y = 3)
-X - 3y = -3

-X - 3y = -3
X + 6y = 3
__________
3y = 0
y = 0/3 = 0.

Now we can solve for x, by simply plugging the value of y into any of the 2 equations.

X + 6y = 3
X + 6(0) = 3
X + 0 = 3
X = 3.

The solution to your system of equations would be (3,0).

Check this by plugging in the point to the other equation and see if it is true.

X + 3y = 3
(3) + 3(0) = 3
3 + 0 = 3
3 = 3.

Thus it is the solution.
5 0
3 years ago
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