Solve the inequality |4x+2|<26

Mathematics

1 answer:

miv72 [106K]3 years ago

7 0

Answer:

-7<x<6

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variables

Mark me brainliest please!

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Help please math thank you

bija089 [108]

Answer:

Step-by-step explanation:

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3 years ago

-(14-2y) +3y+26 how do i find y

Setler79 [48]

Y is behind 2 its in plain sight

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4 years ago

how many gallons of 80% antifreeze solution must be mixed with 100 gallons of 10% antifreeze to get a mixture that is 70% antifr

ololo11 [35]

Let's say we'll mix "x" amount of the 80% solution, how much antifreeze is in it? well is just 80% of antifreeze, and the rest is something else.... what's 80% of "x"? so is (80/100)*x, or 0.8x.

likewise, the mixture will be a 70% solution, and let's say it adds up to "y" amount, so how much antifreeze is in it? well (70/100) * y or 0.7y.

$\bf \begin{array}{lccclll}
&\stackrel{gallons}{amount}&\stackrel{\%}{quantity}&\stackrel{antifreeze}{quantity}\\
&------&------&------\\
\textit{80\% sol'n}&x&0.80&0.8x\\
\textit{10\% sol'n}&100&0.10&10\\
------&------&------&------\\
mixture&y&0.70&0.7y
\end{array}$

so whatever "x" is, we know that x + 100 = y.

and we know that 0.8x + 10 = 0.7y.

$\bf \begin{cases}
x+100=\boxed{y}\\
0.8x+10=0.7y\\
----------\\
0.8x+10=0.7\left( \boxed{x+100} \right)
\end{cases}
\\\\\\
0.8x+10=0.7x+70\implies 0.8x-0.7x=70-10\implies 0.1x=60
\\\\\\
x=\cfrac{60}{0.1}\implies x=600$

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3 years ago

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Lorico [155]

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4 years ago

The formula to find the period of orbit of a satellite around a planet is T^2=(4pi^2/GM)r^3 where r is the orbit's mean radius,

Natalija [7]

The answer is $r= \sqrt[3]{GMT^{2}/4 \pi^{2}}$

$T^{2} = \frac{4 \pi^{2}}{GM} r^{3}$

Move $\frac{4 \pi^{2} }{GM}$ to the other side of the equation:
$T^{2} /\frac{4 \pi^{2} }{GM} = r^{3} \\ 
T^{2} *\frac{GM}{4 \pi^{2} } = r^{3}$

Rearrange:
$r^{3} = T^{2} *\frac{GM}{4 \pi^{2} } \\ 
r^{3}= \frac{T^{2} *GM}{4 \pi^{2} } \\ 
r^{3}= \frac{GMT^{2}}{4 \pi^{2} } \\ 
r^{3} = GMT^{2}/4 \pi^{2}$

Since $x^{3}= \sqrt[3]{x}$ , then
$r^{3} = GMT^{2}/4 \pi^{2} \\ 
r= \sqrt[3]{GMT^{2}/4 \pi^{2}}$

3 0

4 years ago

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