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3 years ago

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An isosceles triangle in which angles B and C have the same measure is shown. Find the measure of each angle whose degree measur

e is represented with variables.
A. x+9y-12
B. 2y+15

C. 4x+15

1 answer:

hoa [83]3 years ago

5 0

hrhhrurgrurhirrurrurrrrrjrbrurbrhtthbt

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3 years ago

Which graph shows the equation c = 10+ 3t, where c is the total cost of going to the carnival and t is the number of $3 tickets

Vladimir [108]

we have

$c = 10+ 3t$

where

c---------> is the total cost of going to the carnival

t---------> is the number of $\$3$ tickets purchased

we know that

<u>the domain of the function is the interva</u>l---------------> [0,∞)

Because the number of tickets can not be negative

<u>The range of the function is the interval</u>-------------> [10,∞)

Because the total cost can not be negative

using a graph tool

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The answer in the attached figure

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Read 2 more answers

If x^3 + 1/x^3 = 110 then find x + 1/x

Svetlanka [38]

$Use:(a+b)^3=a^3+3a^2b+3ab^2+b^3\ (*)\\\\x^3+\left(\dfrac{1}{x}\right)^3\\=\underbrace{x^3+3\cdot x^2\cdot\dfrac{1}{x}+3\cdot x\cdot\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{x}\right)^3}_{(*)}-3\cdot x^2\cdot\dfrac{1}{x}-3\cdot x\cdot\left(\dfrac{1}{x}\right)^2\\\\=\left(x+\dfrac{1}{x}\right)^3-3x-3\cdot\dfrac{1}{x}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)$

therefore

$x^3+\left(\dfrac{1}{x}\right)^3=110\iff \left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)=110\\\\subtitute\ t=x+\dfrac{1}{x}\\\\t^3-3t=110\ \ \ \ |subtract\ 110\ from\ both\ sides\\\\t^3-3t-110=0\\\\t^3-25t+22t-110=0\\\\t(t^2-25)+22(t-5)=0\ \ \ |use\ a^2-b^2=(a-b)(a+b)\\\\t(t-5)(t+5)+22(t-5)=0\\\\(t-5)[t(t+5)+22]=0\iff t-5=0\ or\ \underbrace{t^2+5t+22=0}_{no\ solution}\\\\t=5\iff x+\dfrac{1}{x}=5\\\\Answer:\boxed{x+\dfrac{1}{x}=5}$

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3 years ago