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kifflom [539]
4 years ago
7

Solve using the area model 92×5

Mathematics
1 answer:
shutvik [7]4 years ago
7 0
92x5 is equal to 460
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CAN SOMEONE PLEASE HELP ME!!!
Sindrei [870]

Answer:

2/5, 3/7, and 2/11

Step-by-step explanation:

8 0
3 years ago
Two numbers multiply to be -7 and add to be -6. What are the numbers?<br><br> Example; 2,3
rewona [7]

<u>Given</u>:

Let the two numbers be x and y.

Two numbers multiply to be -7 and add to be -6.

This can be written in equation as,

xy=-7 and

x+y=-6

<u>Value of the two numbers:</u>

Let us determine the value of the two numbers using substitution method.

Substituting y=-6-x in the equation xy=-7, we get;

x(-6-x)=-7

Simplifying, we get;

       -6x-x^2=-7

   x^2+6x-7=0

(x+7)(x-1)=0

x=-7, 1

Thus, the values of x are x = 1,-7

When x = 1 , the equation x+y=-6 becomes y=-7

When x = -7, the equation x+y=-6 becomes y=1

Therefore, the two numbers are 1 and -7

3 0
4 years ago
What are the roots of the function y = 4x2 + 2x – 30? To find the roots of the function, set y = 0. The equation is 0 = 4x2 + 2x
nasty-shy [4]
4x² + 2x - 30 = 0
<span>
factor out the GCF:
</span>2(2x² + x - 15) = 0
<span>
factor the trinomial completely:
2x</span>² + x - 15 = 0
2x² + 6x - 5x - 15 = 0
2x(x + 3) - 5(x + 3) = 0
(2x - 5)(x + 3) = 0

<span>use the zero product property and set each factor equal to zero and solve:
2x - 5 = 0     or     x + 3 = 0
2x = 5                   x = -3
x = 2.5

</span><span>The roots of the function are x=-3,  x=2.5</span>
8 0
3 years ago
Read 2 more answers
-8x+4y=28 solve for y
olga55 [171]
This is how I would do it:
-8x=28-4y divide the entire equation by 4
-2x=7-y Put y back on the left and put x on the right 
y=7-2x.
8 0
4 years ago
Read 2 more answers
If a = -35, b = 10 cm and c = -5, verify that:
Murljashka [212]

Step-by-step explanation:

(i). a+(b+c) = (a+b)+c

-35+(10-5) = (-35+10)+(-5)

-35+5 = -25-5

-30 = -30

(ii). a×(b+c) = a×b + a×c

-35 × [10+(-5)] = -35×10 + -35×-5

-35 × (10-5) = -350 + 175

-35 × 5 = -350 + 175

-175 = -175

7 0
2 years ago
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