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3 years ago

Find the value of the variable and MN if N is between M and Q. Round to the nearest hundredth if necessary.

1 answer:

Rudiy273 years ago

3 0

Answer:
MN = 4 units

Explanation:
1- getting the value of m:
We are given that:
NQ = 4m and NQ = 12
This means that:
12 = 4m
$\frac{4m}{4} = \frac{12}{4}$
m = 3

2- getting MQ:
We are given that:
MQ = 5m + 1
We have calculated that:
m = 3
Therefore:
MQ = 5(3) + 1 = 15 + 1 = 16 units

3- getting MN:
We are given that point N is located somewhere between points M and Q.
This means that:
line MQ can be divided into two portions: MN and NQ
This also means that:
length of MQ = length of MN + length of NQ

We have calculated that:
MQ = 16 units
We are given that:
NQ = 12 units
Therefore:
16 = length of MN + 12
length of MN = 16 - 12
length of MN = 4 units

Hope this helps :)

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3 years ago

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2 years ago

Factor the expression using the two different techniques listed for Parts 1(a) and 1(b).

Natalija [7]

Answer:

9a^4 b^10(2 +3a^6 b^5) (2 -3a^6 b^5)

see work below

Step-by-step explanation:

36a^4b^10 - 81a^16b^20

A) find the GCF

36a^4b^10 = 4*9 a^4b^10 = 2*2*3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

81a^16b^20= 9*9a^16b^20= 3*3*3*3* a*a*a*a*a*a*a*a*a*a*a*a*a*a*a*a *b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b*b

The terms that appear in both terms is the GCF. The terms that remain are inside the parentheses.

The GCF is 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b

36a^4b^10 - 81a^16b^20 = 3*3 a*a*a*a*b*b*b*b*b*b*b*b*b*b *

( 2*2 - 3*3a*a*a*a*a*a*a*a*a*a*a*a b*b*b*b*b*b*b*b*b*b)

Combining like terms

36a^4b^10 - 81a^16b^20 = 9a^4b^10(4-9a^12b^10)

The expression inside the parenthesis can be factored using the difference of squares

let x^2 =4 x =2

y^2 = 9a^12 b^10 y = 3a^6b^5

(x^2 -y^2) = (x+y)(x-y)

9a^4b^10(4-9a^12b^10) = 9a^4b^10 ( 2+3a^6b^5) ( 2-3a^6b^5)

b) difference of squares a^2 – b^2 = (a + b)(a – b)

let a^2 = 36a^4b^10

so a = 6a^2b^5

b^2 = 81a^16b^20

b = 9a^8 b^10

a^2 – b^2 = (a + b)(a – b)

36a^4b^10 - 81a^16b^20 = (6a^2b^5 +9a^8 b^10) (6a^2b^5 -9a^8 b^10)

We can factor a 3 a^2 b^5 out of the first term

3 a^2 b^5 (2 +3a^6 b^5) (6a^2b^5 -9a^8 b^10)

3 a^2 b^5 (2 +3a^6 b^5) 3 a^2 b^5 (2 -3a^6 b^5)

Multiply the terms outside the parentheses together

9a^4 b^10(2 +3a^6 b^5) (2 -3a^6 b^5)

4 0

3 years ago

Read 2 more answers