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In-s [12.5K]
2 years ago
13

How do I factor an expression using the GCF?

Mathematics
1 answer:
Anni [7]2 years ago
4 0

Answer:

  use the distributive property

Step-by-step explanation:

Use the distributive property to rewrite the expression with the GCF factored out.

<u>Example</u>:

  ab +ac +ad  . . . . has terms with a common factor of 'a'

When the common factor is factored out, the distributive property tells you the equivalent is ...

  = a(b +c +d)

If 'a' is the greatest common factor, then b, c, d are mutually prime and no more factors can be removed to outside the parentheses.

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Ramon earns $1,645 each month and pays $53.40 on electricity. To the nearest tenth of a percent, what percent of Ramon’s earning
MAXImum [283]

Answer:

Step-by-step explanation:

54.30/1645 = 0.033 = 3.3%

3 0
3 years ago
There are 15 students in a social studies class. Two students will be selected to present their term projects today. In how many
xz_007 [3.2K]
Given:
15 students
2 students must be chosen.
No repetition, no order

This is a combinations problem. We use this formula: n! / (n-r)!(r!)

n = 15 ; r = 2

15! / (15-2)!(r!) ⇒ 15! / 13! * 2! = 105 
5 0
3 years ago
2x squared plus 16x minus 10 equals 0. please solve for the solution
ohaa [14]
Answer: 5/9

Solution: 2x^2+16x-10=0

According to formula: X=(-16+_sqrt(16^2-(-10•2)) / 2•2

X=5/9
6 0
2 years ago
Read 2 more answers
Factor completely.<br> x² - 3x² - 4
Ber [7]

Answer:

-2(x^2+2)

Step-by-step explanation:

1. x^2-3x^2-4
= -2x^2-4
2. -2(x^2+2) factor completely :)

7 0
2 years ago
Read 2 more answers
When an object is thrown upwards with a speed of 64 ft/sec, its height above the ground is given by the function h(t)=−16t2+64t
Svet_ta [14]

Answer:

The object will reach its highest point 0.5 seconds after it has been thrown.

Step-by-step explanation:

The object reaches its maximum height when velocity is equal to zero, the velocity is the derivative of function height. That is:

v(t) = \frac{d}{dt}h(t) (1)

Where v(t) is the velocity of the object at time t, in feet per seconds.

If we know that h(t) =-16\cdot t^{2}+64\cdot t and v(t) = 0\,\frac{ft}{s}, then the time when object reaches its highest point is:

v(t) = -32\cdot t+64

-32\cdot t + 64 = 0

t = 0.5\,s

The object will reach its highest point 0.5 seconds after it has been thrown.

3 0
2 years ago
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