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Svet_ta [14]
3 years ago
12

Which of the following is a solution to the inequality?

Mathematics
1 answer:
Katena32 [7]3 years ago
8 0

Answer:

a. (0, -3)

Step-by-step explanation:

Put in the inputs and outputs, and see if they come out true or false.

a.

x - 3y ≥ 4

0 - 3(-3) ≥ 4

0 + 9 ≥ 4

9 ≥ 4

True.

b.

x - 3y ≥ 4

0 - 3(5) ≥ 4

0 - 15 ≥ 4

-15 ≥ 4

False.

c.

x - 3y ≥ 4

-4 - 3(-1) ≥ 4

-4 + 3 ≥ 4

-1 ≥ 4

False.

d.

x - 3y ≥ 4

2 - 3(3) ≥ 4

2 - 9 ≥ 4

-7 ≥ 4

False.

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So... our numbers... let's say the first one is hmmm "a"
so the second and subsequent are
a
a+1
a+2
a+3
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there, 5 consecutive whole numbers or integers for that matter

now, we know the sum of the square of the first three,
is the same as the sum of the square of the last two

so \bf \begin{cases}
a\\
a+1\\
a+2\\
\textendash\textendash\textendash\textendash\\
a+3\\
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\end{cases}\qquad (a)^2+(a+1)^2+(a+2)^2=(a+3)^2+(a+4)^2

do a binomial theorem expansion on those, solve for "a"
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3 years ago
Find the similarity ratio of a cube with volume 216 ft to a cube with volume 1000ft
Vika [28.1K]
\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
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\end{array} \\\\
-----------------------------\\\\

\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-----------------------------\\\\
\cfrac{s^3}{s^3}\implies \cfrac{216}{1000}\implies \cfrac{\sqrt[3]{216}}{\sqrt[3]{1000}}\implies \cfrac{s}{s}\impliedby \textit{similarity ratio}

and simplify it away
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3 years ago
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weeeeeb [17]
See diagram
I wrote that the degrees are 70 each because it is isocolese and 140/2=70


a.
to solve for the diagonal, I'm going to use trig to find the height and the base then use pythagorean theorem to find the diagonal

so
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so
we want to find the height and base
sin(70°)=h/7
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cos(70°)=base/7
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b.
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a. 20.68ft
b. 17.21ft

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svp [43]

Answer:

3 to 10 , 3

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There are 3 true questions and 7 false question  , in total , there are 10 questions.

<h3 />
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3 years ago
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