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postnew [5]
3 years ago
11

Please help. I will give brainliest(:

Mathematics
2 answers:
Tanzania [10]3 years ago
7 0
Distribute the -4. Then combine like terms.

24 - 4(5y - 6z) + 3y - 7z =

= 24 - 20y + 24z + 3y - 7z

= - 20y + 3y + 24z - 7z + 24

= -17y + 17z + 24
Musya8 [376]3 years ago
7 0
The answer is D Because you have to add the like terms

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Given: P is a point on the perpendicular bisector, I of MN.<br> Prove: PM = PN
Hatshy [7]

Answer:

Step-by-step explanation:

I can't make specific statements about the proof because the midpoint is missing.

Givens

There are two right angles created by where the perpendicular bisector meats MN. Both are 90 degrees.

MN is bisected by the point on MN where the perpendicular meets MN

The Perpendicular Bisector is is common to both triangles.

Therefore the two triangles are congruent by SAS

PM = PN                Parts contained in Congruent triangles are congruent.

5 0
3 years ago
Point J is on line segment IK. Given JK = 2x – 1, IK = 3x + 2,
Serhud [2]

Answer:

JK=7

Step-by-step explanation:

From the line segment, since J is on it ,it means the line segment is

represented as I J K

from this illustration, we can say that the longest part of the line segment is from I to K

this means that, IJ +JK =IK

making JK the subject,

JK= IK - IJ

but from the question, JK=2x-1 , IK=3x+2 and IJ=3x-5

substituting them in the expression,

2x-1 =3x+2 -(3x-5)

solving for x

2x-1 =3x+2-3x+5

2x-1 =0+7

2x-1 =7

2x=1+7

2x=8

dividing through by 2

2x/2 =8/2

x=4

but the question says we should find the numerical value for JK

but from the line segment,

JK=2x-1

but now we know the value of x to be 2

so substituting it in the formula

JK= 2(4)-1

JK=8-1

JK=7

therefore, the numerical value for JK is 7

4 0
3 years ago
PLEASE HELP MATH
Novosadov [1.4K]
I believe C is your answer. 
I hope this helps!
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3 years ago
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exis [7]

Answer: yellow box

Step-by-step explanation:

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3 years ago
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mina [271]
Is would be a I think because if you times those to numbers you get 20
5 0
3 years ago
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