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melomori [17]
3 years ago
8

PLEASE HELP ME QUICK!!

Mathematics
1 answer:
goldenfox [79]3 years ago
3 0
Supplementary= 180 degrees

It took me a minute- but all you needed to find was half of 180 then divide that by two to get the answer.

180/2= 90

90/2= 45

Angle A= 45(3)= 135

Angle B= 45

135+45= 180

Angle A= 135  Angle B= 45

I hope this helps!
~kaikers
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Solve using quadratic formula<br> 6x^2-x=2
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x = \cfrac{ -(-1) \pm \sqrt { (-1)^2 -4(6)(-2)}}{2(6)}\implies x = \cfrac{1\pm\sqrt{1+48}}{12} \\\\\\ x = \cfrac{1\pm\sqrt{49}}{12}\implies x = \cfrac{1\pm 7}{12}\implies x = \begin{cases} \frac{8}{12}\to &\frac{2}{3}\\[1em] -\frac{6}{12}\to &-\frac{1}{2} \end{cases}

7 0
2 years ago
Patrick paid $12 for<br> 25 pounds of candy.<br> How much did each<br> pound of candy cost?
tiny-mole [99]
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$0.48/pound is your answer
7 0
2 years ago
Read 2 more answers
Read this E[2X^2 â€" Y].
djyliett [7]

Looks like a badly encoded/decoded symbol. It's supposed to be a minus sign, so you're asked to find the expectation of 2<em>X </em>² - <em>Y</em>.

If you don't know how <em>X</em> or <em>Y</em> are distributed, but you know E[<em>X</em> ²] and E[<em>Y</em>], then it's as simple as distributing the expectation over the sum:

E[2<em>X </em>² - <em>Y</em>] = 2 E[<em>X </em>²] - E[<em>Y</em>]

Or, if you're given the expectation and variance of <em>X</em>, you have

Var[<em>X</em>] = E[<em>X</em> ²] - E[<em>X</em>]²

→   E[2<em>X </em>² - <em>Y</em>] = 2 (Var[<em>X</em>] + E[<em>X</em>]²) - E[<em>Y</em>]

Otherwise, you may be given the density function, or joint density, in which case you can determine the expectations by computing an integral or sum.

6 0
2 years ago
The value of y is directly proportional to the value of x If y=81 when x=54, what is the value of y when x=38
Arisa [49]

Answer:

y = 57

Step-by-step explanation:

y α x

y = kx ; k = Constant of proportionality

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y = kx

y = 1.5 * 38

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Hence, y = 57 when x = 38

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2 years ago
Need Help! 10 points!
Dominik [7]
The answer is d because 4+8=12 and they are 4 units away from eachother:)
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