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hichkok12 [17]
3 years ago
5

Multiplying a trinomial by a trinomial follows the same steps as multiplying a binomial by a trinomial. Determine the degree and

maximum possible number of terms for the product of these trinomials: (x2 + x + 2)(x2 – 2x + 3). Explain how you arrived at your answer.

Mathematics
1 answer:
GrogVix [38]3 years ago
6 0
I think this attachment will help you

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-5(5c-3b-1) distribute property
andrew11 [14]

Answer:-25c+15b+5 ..add it up then boom you'll get your answer

Step-by-step explanation:

4 0
3 years ago
15x+5y=10 <br>5x-7y=12<br><br>solve using elimination ​
Paul [167]
Point form: (1,-1)
equation form: x = 1, y = -1
8 0
3 years ago
Sharon was going through the financial records of her company. The profit earned by the company, p(t), over time t, in years, fo
FromTheMoon [43]
For this case what we must do is find a quadratic function that is already factored.
 This is because in the factored quadratic equations, it is easier to observe the zeros of the function.
 In this case, the zeros of the function represent the time at which the company did not make any profit.
 We have the following equation:
 p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
 We observed that there was no gain in:
 t = 3
 t = 5
 The other roots are discarded because they are negative
 Answer:
 
a.p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
5 0
3 years ago
A trapezoid has a perimeter of 14 feet. What will the perimeter be if each side length is increased by a factor 7? 21 feet 98 fe
natka813 [3]

Answer:

Correct choice is B

Step-by-step explanation:

Let a, b, c and d be the trapezoid's sides lengths.

The perimeter of the trapezoid is the sum of all sides lengths, thus,

a+b+c+d=14\ ft.

If each side length is increased by a factor 7, then new sides have lengths 7a, 7b, 7c and 7d. The perimeter of new trapezoid is

7a+7b+7c+7d.

Use the distributive property for this expression:

7a+7b+7c+7d=7(a+b+c+d).

Since a+b+c+d=14\ ft, then

7a+7b+7c+7d=7(a+b+c+d)=7\cdot 14=98\ ft.

7 0
3 years ago
Read 2 more answers
The distance from Joshua's house to the mall is 3.5 miles. The distance from Meg's house to the mall is 19,000 feet. Who lives c
damaskus [11]

Answer:

i think its Joshua

7 0
3 years ago
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