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igor_vitrenko [27]
3 years ago
6

Please HELP me lol I need it !!

Mathematics
1 answer:
Xelga [282]3 years ago
6 0
Graph D.
The interval goes from -1 to 4. -1 is included so it'll have a filled circle. 4 isn't included so it'll have a hollow circle.
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Joyce paid $60.50 for an item at the store that was 45 percent off the original price. What was the original price?
Lynna [10]

Answer:

27.23

Step-by-step explanation:

60.50 divided by .45

4 0
2 years ago
Can someone help me solve this question I can't seem to manage it
MAXImum [283]

Answer:

LA=6 WA=4 LB=6 WB=2

Step-by-step explanation:

Its a square so the sides are 6cm. 2:1 ratio means A is 6*4=24 cm while B is 6*2=12cm.

6 0
3 years ago
ΔMNP is an isosceles triangle with MN ≈NP. If ∠ M = 3x + 1 and ∠ N = x - 11. What is the measure of ∠P?
s2008m [1.1K]

Start with an equation summing all the angles in this triangle:

180 = <M + <N + <P

we are given <M and <N but not <P. But, since MN=NP, the angle <P is the same as the angle <M (isosceles, make a drawing to see). So

180 = 2<M + <N

180 = 2(3x+1) + x-11

180 = 7x - 9

x = 27

<P = 3*27+1 = 82 degrees

4 0
3 years ago
X= -5 y=2 work out the VALUE if 3x + 4y
Orlov [11]
3(-5) + 4(2)
= -15 + 8
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7 0
2 years ago
Read 2 more answers
Solve the following system
scZoUnD [109]

Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

4 0
3 years ago
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