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3 years ago

I'm desperate for help. Please help

Mathematics

2 answers:

Rus_ich [418]3 years ago

7 0

Remember that in PEMDAS, you do the parentheses first.

1. For each answer choice, do whatever is in the parentheses first.

2. Multiply that number to the one outside the parentheses

Hint: For this problem, If all the signs in the answer choice are multiplication, they will equal the same number.

alukav5142 [94]3 years ago

5 0

It would be a!

step by step : in PEMDAS you would completer the parentheses first. 10*7 =70 * 4 =280. 10*4=40 *7 also = 280 hope that helps!!

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In the diagram below of triangle UVW, X is a midpoint of UV and Y is a midpoint of VW. If mZY XV = 3x + 48, and mZWUX = x + 62,

nika2105 [10]

Given:

m∠YXV = 3x + 48

m∠WUX = x + 62

Using the corresponding angle theorem, since X is the midpoint of UV and Y is the midpoint VW, m∠YXV is congruent to m∠WUX.

Thus, we have:

m∠YXV = m∠WUX

3x + 48 = x + 62

Solve for x.

Subtract x from both sides:

3x - x + 48 = x - x + 62

2x + 48 = 62

Subtract 48 from both sides:

2x + 48 - 48 = 62 - 48

2x = 14

Divide both sides by 2:

$\begin{gathered} \frac{2x}{2}=\frac{14}{2} \\ \\ x=7 \end{gathered}$

To find m∠WUX, where x = 7, we have:

m∠WUX = x + 62

= 7 + 62

= 69º

ANSWER:

m∠WUX = 69º

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2 years ago

If DE bisects <ABC.find the value of x if m<ABC=7x+29 and m<DBC=99°

SashulF [63]

Answer:

Step-by-step explanation:

If DE bisects <ABC, this means that <ABE = <EBC

Also <DBC+<EBC = 180

99+<EBC = 180

<EBC = 180-99

<EBC = 81

Also <ABC =<ABE+<EBC

<ABC =<EBC+<EBC

7x+29 = 81+81

7x = 162-29

7x = 133

x = 133/7

x = 19°

Hence the value of x is 19°

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3 years ago

Find the TWO integers whos product is -12 and whose sum is 1<br>

Deffense [45]

Answer:

$\rm Numbers = 4 \ and \ -3.$

Step-by-step explanation:

Given :-

The sum of two numbers is 1 .

The product of the nos . is 12 .

And we need to find out the numbers. So let us take ,

First number be x

Second number be 1-x .

According to first condition :-

$\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}$