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1 year ago

Use prime factors to determine the HFC of 90 and 126

Mathematics

2 answers:

Gala2k [10]1 year ago

6 0

Highest common prime factor is three

ohaa [14]1 year ago

6 0

Answer:

Step-by-step explanation:

Comment

The question means that you are to break both numbers down in to prime factors. When you do, what do the numbers have in common

90: 2 * 3 * 3 * 5

126: 2 * 3 * 3 * 7

The bolded numbers are the primes in common

Primes in common (answer): 2 * 3 * 3 = 18

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Some one please help me with this question

pychu [463]

The answer is C because you do PEMDAS and then add the 5 from the other side

5 0

2 years ago

Solve the inequality. 9 + 9(x + 7) >= 36 The solution is

Alchen [17]

Answer:

x≥ -4

Step-by-step explanation:

9 + 9(x + 7)≥ 36

Subtract 9 from each side

9 -9+ 9(x + 7) ≥ 36-9

9(x + 7) ≥ 27

Divide each side by 9

9/9(x + 7) ≥ 27/9

x+7≥ 3

Subtract 7 from each side

x+7-7 ≥ 3-7

x≥ -4

6 0

2 years ago

Read 2 more answers

Suppose the top surface of a balance beam is 16 feet 5 inches long and 3.9 inches wide. Find the area of the top surface of the

VARVARA [1.3K]

Answer:

Option A would be your answer

Step-by-step explanation:

Had this question on Achieve3000

7 0

2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

2 years ago

Lamont lost 8 yards in the first quarter

beks73 [17]

The answer is -6 sorry could not explain because have to go somewhere hope it helped

3 0

2 years ago

Use prime factors to determine the HFC of 90 and 126​

Use prime factors to determine the HFC of 90 and 126