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olga_2 [115]
3 years ago
13

Solve each system of equations algebraically.

Mathematics
1 answer:
brilliants [131]3 years ago
6 0
(0,0) hope this helps!
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 ≥ can be written in both the circles

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3 years ago
A rectangle has sides measuring (6x + 4) units and (2x + 11) units. Part A: What is the expression that represents the area of t
valentinak56 [21]

Answer:

Step-by-step explanation:

Part A:

A=lw

A=(6x + 4)*(2x + 11)

multiply it out:

A=6x*2x+6x*11+4*2x+4*11

A=12x^(2)+66x+8x+44

add like terms

A=12x^(2)+<u>66x+8x</u>+44

A=12x^(2)+(66+8)x+44

A=12x^(2)+74x+44

the expression used to find area is 12x^(2)+74x+44

Part B:

this is:

a second degree trinomial

since its leading coefficient (highest degree value of the expression) is 2

and

since it has three monomials (terms that are separated by + or - signs):

12x^(2)+<u><em>74x</em></u>+44

Part C:

(not sure about this one)

all area is positive or 0 (no area), so lets find the values of x that are true:

lets find the zeros

(6x + 4)*(2x + 11)=0

6x+4=0

6x=-4

x=-4/6

x=-2/3

2x+11=0

2x=-11

x=-11/2

6*(-11/2)+4

3*(-11)+4

-33+4

-29

2*(-2/3)+11

-4/3+33/3

29/3

6*(-2/3) + 4

2*(-2)+4

-4+4

0

x is greater than (-2/3)

Thus, any real number greater than -2/3 makes

(6x + 4)*(2x + 11) equals the expression:

12x^(2)+74x+44, where x is a real number greater than -2/3

if lets say x= 1, a counting number then:

(6*1 + 4)*(2*1 + 11)=12*1^(2)+74*1+44

(6+4)*(2+11)=12*1+74+44

10*13=12+118

130=130

130 and 1 are both counting numbers, proving the closure property of polynomials

7 0
3 years ago
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What’s the solution of 5(2m - 13) =65
earnstyle [38]

Answer:

m=13

Step-by-step explanation:

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3 years ago
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What non-zero rational number must be placed in the square so that the simplified product of these two binomials is a binomial:
aleksandrvk [35]

The product would expand to

(7t-10)(5t+x)=35t^2+7tx-50t-10x=35t^2+(7x-50)t-10x

This is a trinomial, and the only way to make it a binomial is to cancel out a coefficient using our variable x.

So, we can cancel either the linear term or the constant term.

In the first case, we require

7x-50=0 \iff 7x=50 \iff x=\dfrac{50}{7}=\dfrac{49}{7}+\dfrac{1}{7}=7\dfrac{1}{7}

In the second case, we require

-10x=0\iff 10x=0 \iff x=0

But x must be a non-zero rational number, so this solution is not feasible.

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Answer:

100%

Step-by-step explanation:

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