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vivado [14]
4 years ago
7

Which pair of numbers has a GCF of 3? (A) 1 and 6 (B) 2 and 3 (C) 2 and 6 (D) 3 and 4

Mathematics
1 answer:
Elden [556K]4 years ago
7 0

Answer:

None of the above

Step-by-step explanation:

To find the GCF of something find the prime factorization of each number, then see which numbers are the same. After that multiply the numbers that are the same.

(PF): Prime Factorization

A: PF - 1   PF - 2 · 3 (Since they don't have any numbers in common the GCF is 1)

B: PF - 2  PF - 3 (The GCF is also 1)

C: PF - 2  PF - 2 · 3 (Because they have 2 in common the GCF is 2)

D: PF - 3  PF - 2 & 2 (The GCF is 1)

None of the options have a GCF of 3. Therefore it is none of the above

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Zielflug [23.3K]
The easiest way to do this is to multiply 12km by 1.5. That will give you one and half the value of 12, which will represent the actual distance between the two cities.
12 * 1.5 = 18
The two cities are actually 18 kilometers away from each other.
6 0
3 years ago
What is the general term equation, an, for the arithmetic sequence 9, 5, 1, −3, . . . , and what is the 21st term of this sequen
marin [14]
Hello : 
<span>the general term equation, an, for the arithmetic sequence is : 
An = A1+d(n-1)
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4 0
3 years ago
Nick and Adam workout together. Nick weighs 160 pounds and is gaining about 3 pounds per week. Adam weighs 195 pounds and is los
Gekata [30.6K]

Answer:

x=7

Step-by-step explanation:

(160+3x)=(195-2x)

4 0
3 years ago
Pierre has more money than alex.
Lelechka [254]

Answer:

p = 2a-22

a= p-20

these will help you find the actual value

Step-by-step explanation:

this is not for the actual value

pierre = p

alex = a

p = a+20

p-20 = a

p+22 = 2a

p = 2a-22

7 0
3 years ago
The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width
Kay [80]

Answer:

The length of the rectangle is 11 inch and its width is 8 inch.

Step-by-step explanation:

Given:

Rectangle (R_1) and rectangle (R_2).

Let the width of the rectangle be 'x' inch.

Its length = '(x+3)' inch

Area of the rectangle = (A_1)

⇒A_1 = (x)(x+3)

Now for rectangle (R_2) its width is decreased by 1 inch and length is doubled,the area (A_2) will increased by 66 sq-inch.

So,

A_2=A_1+66

According to the question.

⇒ Area of rectangle (R_2)  with width '(x-1)' and length '2(x+3)'='(2x+6)' .

⇒ A_2=(x-1)(2x+6)

Plugging the values of A_2 in terms of A_1.

⇒ A_1+66= (x-1)(2x+6)

Plugging  A_1 = (x)(x+3) into the equation.

⇒ (x)(x+3)+66=(x-1)(2x+6)

⇒ x^2+3x+66=2x^2+6x-2x-6

⇒ x^2-2x^2+3x-6x+2x+6+66=0

⇒ -x^2-x+72=0

Now using quadratic formula we can find the value of 'x'.

Quadratic formula, (x)= \frac{\pm b+\sqrt{b^2-4ac} }{2a}

x= -9 and x= 8

Discarding the negative value.

Our width  = 'x' = 8 inches.

So the length of the rectangle = 11 inches and width of the rectangle = 8 inches.

7 0
3 years ago
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