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3 years ago

(64-2^2)-(8+7)answer please

1 answer:

7 0

Hello there!

64 - 2² - (8 + 7)

= 64 - 4 - 15

= 60 - 15

= 45

Let me know if you have any misunderstanding!

As always, I am here to help!

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Kyle volunteered 5.5 hours at the library this week. He needs to volunteer a total of 12 hours. Which equation can be used to fi

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12-5.5=h hope it helps

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3 years ago

Please help asap! this isn’t too hard but i am still confused.

juin [17]

Answer:

The consecutive even numbers are 28, 30, 32, 34

Step-by-step explanation:

let the four consecutive even numbers be: b d f h,

such that:

b=b, d=b+2, f=b+4, h=b+6.

Therefore, 2(b+b+2)=3(b+6)+14

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3 years ago

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Which variable expression represents the phrase “the quotient of 5 times a number and 2”?

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5x/2

Step-by-step explanation:

Say the mystery number is "x"

5 times that = 5 * x = 5x

Then the quotient (that means the result of division, so divide!) of that and 2 would give you: 5x/2

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2 years ago

What is the equation of a circle with a center at (–4, 0) that passes through the point (–2, 1)? x2 + (y + 4)² = StartRoot 5 End

Allisa [31]

Answer:

Step-by-step explanation:

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3 years ago

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How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

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$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

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$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$

Neither solution makes any denominator be zero, so both are good solutions.

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3 years ago