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IceJOKER [234]
2 years ago
13

How do you solve this math problem using substitution n=5n n=2/3m-13

Mathematics
1 answer:
Illusion [34]2 years ago
5 0

Answer:

I love Math anyways  

the ans is in the picture with the steps  

(hope it helps can i plz have brainlist :D hehe)

Step-by-step explanation:

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I will mark brainliest if you answer correctly! Please! I need this ASAP!!!
Nikitich [7]
C. To find the y-intercept, substitute in 0 for x and solve for y. y-intercept= (0,33)
4 0
3 years ago
1/10 is one part of the whole(number)divided into.......equals part.
Tom [10]
10 equal parts. that is the answer
5 0
3 years ago
Help please!!!! i dont know how to do this
IRINA_888 [86]

All the angles created by the transversal intersecting through a pair of parallel lines have got many names and connections with each other, like Alternate Angles, Corresponding angles, consecutive interior angles etc.

As per the question statement, We are given a pair of parallel lines which is cut by a transversal. We are supposed to mark the following angles.

Alternate Interior Angles, Alternate Exterior Angles, Corresponding Angles and Consecutive Interior Angles.

Here is an attached image of the same with angles marked on it.

Alternate Interior Angles: ∠3 = ∠5 and ∠4=∠6

Alternate Exterior Angles: ∠2=∠8 and ∠1=∠7

Corresponding Angles: ∠1=∠5 , ∠4=∠8 , ∠3=∠7 and ∠2=∠6

Consecutive Interior Angles: ∠4=∠5 and ∠3=∠6

  • Parallel Lines: Parallel lines are those straight lines that are, no matter how far they are extended, always the same distance apart from one another.
  • Transversal Line: In geometry, a transversal line intersects two lines in the same plane at two different locations.

To learn more about Transversal Line click on the link given below:

brainly.com/question/24770636

#SPJ1

8 0
11 months ago
How many turning points can a polynomial with a degree of 7 have?
aleksklad [387]
That would be 7 - 1  = 6 turning points
5 0
3 years ago
Part A: Factor x2y2 + 6xy2 + 8y2. Show your work. (4 points) Part B: Factor x2 + 8x + 16. Show your work. (3 points) Part C: Fac
skad [1K]

Answer:

Part A : y²(x + 2)(x + 4)

Part B: (x + 4) (x + 4)

Part C: (x + 4) (x - 4)

Step-by-step explanation:

Part A: Factor x²y²+ 6xy²+ 8y²

x²y²+ 6xy²+ 8y²

y² is very common across the quadratic equation , hence

= y² (x² + 6x + 8)

= (y²) (x² + 6x + 8)

= (y²) (x² + 2x +4x + 8)

= (y²) (x² + 2x)+(4x + 8)

= (y²) (x(x + 2)+ 4(x + 2))

= y²(x+2)(x+4)

Part B: Factor x² + 8x + 16

x² + 8x + 16

= x² + 4x + 4x + 16

= (x² + 4x) + (4x + 16)

= x( x + 4) + 4(x + 4)

= (x + 4) (x + 4)

Part C: Factor x² − 16

= x² − 16

= x² + 0x − 16

= x² + 4x - 4x - 16

= (x² + 4x) - (4x - 16)

= x (x + 4) - 4(x + 4)

= (x + 4) (x - 4)

6 0
3 years ago
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