Find two consecutive even integers whose product is 288

6 answers
Question:

Find two consecutive even integers whose product is 288

Answers

So I’m going to solve this using algebra OK?
So what you do is take the first integer as X.
Then we know that x(the first number) + X+2(the second number) is 288 right?
So we solve that saying x+x+2=288.
So 2X+2=288
2X+2-2=288-2 right?
So 2X=286.
So x is 286/2 right?
So x is 143.
Now the first number is 143 right?
So the second number should be 143+2 so 145.
I know they are not even but these are the solutions.
I don’t think there are any consecutive even numbers that add up to 288

(X)+(X+1)=24

"consecutive" means right after the other, so, if the first one is (X), the second integer must be (X+1). The answer 24 is given so you just have to solve the equation.

2X+1=24
2X=23
X=11.5

10 and 12 are two even integers whose product =120

Step-by-step explanation:

10 and 12 are two even integers whose product =120

[tex]10\times 12=120[/tex]

solve the answer

Step-by-step explanation:

let the two integers be x and x+1 as they are consecutive

so x(x+1) = 24

x^2 + x = 24

and now solve the quadratic equatioon

QUESTION 1

GIVEN:

[tex]long \: base = 3 \times (short \: base)[/tex]

[tex]height = (short \: base) - 2[/tex]

[tex]area = 30 {in}^{2}[/tex]

IMPORTANT EQUATIONS

[tex]trapezoid \: area = \frac{sum \: of \: bases}{2} \times height[/tex]

SOLVE:

[tex]30 = \frac{short + long}{2} (short - 2)[/tex]
[tex]30 = \frac{(short + 3short)}{2} \times (short - 2)[/tex]

[tex]30 = \frac{4short}{2} (short - 2)[/tex]

[tex]30 = 2{(short)}^{2} - 4(short)[/tex]

[tex]{short}^{2} - 2(short) - 15 = 0[/tex]

factor:

[tex](short - 5)(short + 3) = 0[/tex]

[tex]short = 3 \: and \: - 5[/tex]
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

[tex]short \: base = 3 \: in[/tex]

[tex]long \: base = 3 \times 3 = 9 \: in[/tex]

[tex]height = 3 - 2 = 1 \: in[/tex]

QUESTION 2

GIVEN
convert 3ft to inches.

[tex]l = 32 + w[/tex]

[tex]area = lw = 130 {in}^{2}[/tex]
or

[tex]w = \frac{130}{l}[/tex]

substitute

[tex]l = 32 + \frac{130}{l}[/tex]

solve for l:

[tex]{l}^{2} - 32 l- 130 = 0[/tex]
QUESTION 3

[tex]x(x + 2) = 80[/tex]

solve for X

[tex]{x}^{2} + 2x - 80 = 0[/tex]
[tex](x - 10)(x + 8) = 0[/tex]

[tex]x = 10 \: and \: - 8[/tex]

take the positive value since it's the only one that makes sense in this context.

so the two values are:

[tex]10 \: and \: 8[/tex]

You can't find the precise integer without knowing what the product is - your question is incomplete - but you can find a general solution.

If [tex]x[/tex] is one of the integers, then the next even integer is given by [tex]x+2[/tex]. So if the product is [tex]n[/tex], then you have

[tex]x(x+2)=k\implies x^2+2x=k\implies x^2+2x+1=k+1\implies (x+1)^2=k+1\implies x=-1\pm\sqrt{k+1}[/tex]

Plug in [tex]k[/tex] and select the solution for [tex]x[/tex] that is an even integer.

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