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3 years ago

Use prime factorization to find the GCF of 140 and 180

Mathematics

2 answers:

Ierofanga [76]3 years ago

8 0

Answer:

Step-by-step explanation:

Kisachek [45]3 years ago

7 0

Answer:

Step-by-step explanation:

Common Factor (GCF) is: 2 x 2 x 5 = 20

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I’m stuck can some one help me

abruzzese [7]

Answer:

3) Midpoint is (-4,0.5)

Option A is correct.

4) Midpoint is (2.5,0)

Option B is correct.

5) The factors are (x+4)(x-7)

Option C is correct.

6) The factors are (x+4)(x+2)

Option A is correct.

Step-by-step explanation:

Question 3

Find midpoint of the following:

(2,-7), (-10,8)

The formula used to find midpoint is: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

We have $x_1=2, y_1=-7, x_2=-10,y_2=8$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2-10}{2},\frac{-7+8}{2} )\\Midpoint=(\frac{-8}{2},\frac{1}{2} )\\Midpoint=(-4,0.5 )$

So, Midpoint is (-4,0.5)

Option A is correct.

Question 4

Find midpoint of the following:

(2,-10), (3,10)

The formula used to find midpoint is: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

We have $x_1=2, y_1=-10, x_2=3,y_2=10$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )\\Midpoint=(\frac{2+3}{2},\frac{-10+10}{2} )\\Midpoint=(\frac{5}{2},\frac{0}{2} )\\Midpoint=(2.5,0 )$

So, Midpoint is (2.5,0)

Option B is correct.

Question 5

Factor each completely

$x^2-3x-28$

We will break the middle term and find factors

$x^2-3x-28\\=x^2-7x+4x-28\\Taking\:common\\=x(x-7)+4(x-7)\\=(x+4)(x-7)$

So, the factors are (x+4)(x-7)

Option C is correct.

Question 6

Factor each completely

$x^2+6x+8$

We will break the middle term and find factors

$x^2+6x+8\\=x^2+4x+2x+8\\=x(x+4)+2(x+4)\\=(x+4)(x+2)$

So, the factors are (x+4)(x+2)

Option A is correct.

7 0

3 years ago

√48a^4 b^5 c^3 d simplify and keep it in radical form

Morgarella [4.7K]

I will assume the square root extends all the way across.

$\sqrt{48a^4b^5c^3}$

$\sqrt{2^4\cdot3\cdot a^4b^5c^3}$

$\sqrt{(2^2)^2\cdot3\cdot(a^2)^2\cdot(b^2)^2\cdot b\cdot c^2\cdot c}$

$\sqrt{(4)^2\cdot3\cdot(a^2)^2\cdot(b^2)^2\cdot b\cdot c^2\cdot c}$

$\sqrt{(4\cdot a^2 \cdot b^2 \cdot c)^2\cdot3\cdot b\cdot c}$

$4\cdot a^2\cdot b^2\cdot c\sqrt{3\cdot b\cdot c}$

$\boxed{4a^2b^2c\sqrt{3bc}}$

3 0

3 years ago

(d) Given: AB = 4; in and AC = 3<br> Find: The length of BC or BC<br> What the answer??

mars1129 [50]

Your total answer is 7.

8 0

3 years ago

The two triangles are similar What is the value of x? Enter your answer in the box x =

lilavasa [31]

Answ x=59

Step-by-step explanation:

4 0

2 years ago

A 747 airplane weighs about 600,000 pounds. It can also be expressed as approximately 3 times 10 Superscript n tons. If there ar

miss Akunina [59]

Answer:

n=2

Step-by-step explanation:

2000 pounds = 1 ton 600000 pounds = X tons Hence X tons = (600000 x 1) ÷ 2000 = 300 tons 3 x 10ⁿ = 300 Hence 3 x 10ⁿ = 3 x 10² Hence n= 2

5 0

3 years ago

Read 2 more answers