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2 years ago

Can anyone answer this question? I'll be giving 100 points, and mark brainliest to the best answer!!

Mathematics

2 answers:

Temka [501]2 years ago

8 0

Answer:

see below

Step-by-step explanation:

<APR and < PRD are alternate interior angles, when lines APB and CRD are parallel, alternate interior angles are equal

<APR = < PRD

62+x =115

Subtract 62 from each side

62+x-62 = 115-62

x =53

nasty-shy [4]2 years ago

4 0

X=53 here is your answer and plsss mark me as a brainliest

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If f(x) = 53 - 2x, then f(-3) =

zheka24 [161]

Substitute -3 for x in the expression

f(x)= 53 - 2x

f(-3)= 53 - 2(-3)
multiply -2 * -3

f(-3)= 53 + 6
add

f(-3)= 59

ANSWER: f(-3)= 59

Hope this helps! :)

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3 years ago

If i got 17 questions out of 20 correct then what is my grade on the paper?

Svetradugi [14.3K]

Hello,

Your answer would be:

75%

Plz mark me brainliest

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3 years ago

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How do you express equivalent equation

ExtremeBDS [4]

Http://www.icoachmath.com/math_dictionary/equivalent_equations.html

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3 years ago

0.7x+6=0.3(4x+6)−2x

Gelneren [198K]

Answer: $x=-2.8$

Step-by-step explanation:

Given the following equation:

$0.7x+6=0.3(4x+6)-2x$

You can follow these steps in order to solve for "x" and find its value:

1. Apply Distributive property on the right side of the equation:

$0.7x+6=0.3(4x+6)-2x\\\\0.7x+6=1.2x+1.8-2x$

2. Subract $0.7x$ from both sides and add the like terms:

$0.7x+6-(0.7x)=1.2x+1.8-2x-(0.7x)\\\\6=-1.5x+1.8$

3. Subtract 1.8 from both sides:

$6-(1.8)=-1.5x+1.8-(1.8)\\\\4.2=-1.5x$

4. Finally, divide both side of the equation by -1.5:

$\frac{4.2}{-1.5}=\frac{-1.5x}{-1.5}\\\\x=-2.8$

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4 years ago

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Which expression is equivalent to StartFraction b Superscript negative 2 Baseline Over a b Superscript negative 3 Baseline EndFr

nignag [31]

Answer:

$\frac{b^{-2}}{ab^{-3}}=\frac{b}{a}$

Step-by-step explanation:

Writing the description in algebraic translation

$\frac{b^{-2}}{ab^{-3}}$

so we have to find the expression which will be equal to $\frac{b^{-2}}{ab^{-3}}$ .

Considering the expression

$\frac{b^{-2}}{ab^{-3}}$

$\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}$

$\frac{b^{-2}}{b^{-3}}=b^{-2-\left(-3\right)}$

so the expression becomes

$=\frac{b^{-2-\left(-3\right)}}{a}$ ∵ $\frac{b^{-2}}{b^{-3}}=b^{-2-\left(-3\right)}$

$\mathrm{Subtract\:the\:numbers:}\:-2-\left(-3\right)=1$

$=\frac{b}{a}$

Therefore,

$\frac{b^{-2}}{ab^{-3}}=\frac{b}{a}$

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4 years ago

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