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Kazeer [188]
3 years ago
5

How are the factors of 36 related to the factors of 72

Mathematics
1 answer:
Zolol [24]3 years ago
8 0
They are related because some of the factors of 36 overlap with 72
EX: 36 - 1,2,3,4,6,12,18
EX: 72 - 1,2,3,4,6,12,18,31

So that means 1,2,3,4,6,12,18 overlap
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Find value of x, show your work
Zina [86]
Here is your answer

\huge x=\frac{35}{2}cm= 17.5cm

REASON :
AB//CD

So,

/_A=/_D ... (alternate int. angles)

/_B=/_C ... (alt. int. angles)

So,

Tri.ABE~Tri.DCE... (by AA similarity)

Therefore,

Corresponding sides are proportional

i.e.

\frac{AE}{DE}=\frac{AB}{DC}

\frac{5}{x}=\frac{4}{14}

x=\frac{5×14}{4}

x=\frac{5×7}{2}

x=\frac{35}{2}

HOPE IT IS USEFUL
5 0
3 years ago
Write one hundred eleven in expanded form
nydimaria [60]
Answer: 100+10+1

You basically place zeroes instead of the actual digit to find the value of each number.

Example:

855

800+50+5
6 0
3 years ago
What’s the answer for this question?
qaws [65]

Answer:

d.

Step-by-step explanation:

you would put 4 in the x places and 2 in the yplaces and then you would distribute 4^2 +3(2)/4(2)

16+6/2=22/8=2 3/4

8 0
3 years ago
If x = (√2 + 1)^-1/3 then the value of x^3 + 1/x^3 is​
Shtirlitz [24]

Step-by-step explanation:

<u>Given</u><u>:</u> x = {√(2) + 1}^(-1/3)

<u>Asked</u><u>:</u> x³+(1/x³) = ?

<u>Solution</u><u>:</u>

We have, x = {√(2) + 1}^(-1/3)

⇛x = [1/{√(2) + 1}^(1/3)]

[since, (a⁻ⁿ = 1/aⁿ)]

Cubing on both sides, then

⇛(x)³ = [1{/√(2) + 1}^(1/3)]³

⇛(x)³ = [(1)³/{√(2) + 1}^(1/3 *3)]

⇛(x)³ = [(1)³/{√(2) + 1}^(1*3/3)]

⇛(x)³ = [(1)³/{√(2) + 1}^(3/3)]

⇛(x * x * x) = [(1*1*1)/{√(2) + 1)^1]

⇛x³ = [1/{√(2) + 1}]

Here, we see that on RHS, the denominator is √(2)+1. We know that the rationalising factor of √(a)+b = √(a)-b. Therefore, the rationalising factor of √(2)+1 = √(2) - 1. On rationalising the denominator them

⇛x³ = [1/{√(2) + 1}] * [{√(2) - 1}/{√(2) - 1}]

⇛x³ = [1{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Multiply the numerator with number outside of the bracket with numbers on the bracket.

⇛x³ = [{√(2) + 1}/{√(2) + 1}{√(2) - 1}]

Now, Comparing the denominator on RHS with (a+b)(a-b), we get

  • a = √2
  • b = 1

Using identity (a+b)(a-b) = a² - b², we get

⇛x³ = [{√(2) - 1}/{√(2)² - (1)²}]

⇛x³ = [{√(2) - 1}/{√(2*2) - (1*1)}]

⇛x³ = [{√(2) - 1}/(2-1)]

⇛x³ = [{√(2) - 1}/1]

Therefore, x³ = √(2) - 1 → → →Eqn(1)

Now, 1/x³ = [1/{√(2) - 1]

⇛1/x³ = [1/{√(2) - 1] * [{√(2) + 1}/{√(2) + 1}]

⇛1/x³ = [1{√(2) + 1}/{√(2) - 1}{√(2) + 1}]

⇛1/x³ = {√(2) + 1}/[{√(2) - 1}{√(2) + 1}]

⇛1/x³ = [{√(2) + 1}/{√(2)² - (1)²}]

⇛1/x³ = [{√(2) + 1}/{√(2*2) - (1*1)}]

⇛1/x³ = [{√(2) + 1}/(2-1)]

⇛1/x³ = [{√(2) + 1}/1]

Therefore, 1/x³ = √(2) + 1 → → →Eqn(2)

On adding equation (1) and equation (2), we get

x³ + (1/x³) = √(2) -1 + √(2) + 1

Cancel out -1 and 1 on RHS.

⇛x³ + (1/x³) = √(2) + √(2)

⇛x³ + (1/x³) = 2

Therefore, x³ + (1/x³) = 2

<u>Answer</u><u>:</u> Hence, the required value of x³ + (1/x³) is 2.

Please let me know if you have any other questions.

3 0
2 years ago
I need this answered please
TiliK225 [7]
The answers are

1: False
2: True
3: True
4: False
4 0
2 years ago
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