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3 years ago

Please answer this question, and explain it to me as well how you got the answer.

1 answer:

5 0

You can look for a factor of a 3 or 5 within the values 16 and 24.
24 can be factored as 3*8.
So you can break up the 24 in that way before multiplying,
and then have the 3's divide evenly out,
leaving you with a simpler problem.
$\rm \frac3{16}\cdot\frac5{24}\quad=\quad \frac{3\cdot5}{16\cdot24}\quad=\quad \frac{3\cdot5}{16\cdot3\cdot8}\quad=\quad \frac{5}{16\cdot8}$

Is this what you were looking for perhaps?

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SCORPION-xisa [38]

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3 years ago

Divide (x^3+2x^2-x-2) by (x-1)

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3 years ago

(Can someone help me please?)

Scorpion4ik [409]

Answer:

Step-by-step explanation:

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3 years ago

9•(-5) A. -45 B. -95 C. 4 D. 45

alexgriva [62]

Answer:

A. -45

Step-by-step explanation:

multiplying a positive an a negative number gives out a negative result

so when we multiple 9 with -5 the result is -45

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3 years ago

Read 2 more answers

If the original square had a side length of

irina [24]

Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$