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2 years ago

9

Solve the inequation; 5x²+82>262

1 answer:

eduard2 years ago

3 0

5x^2+82>262
-82. -82
5x^2>180
5x^2/5>180/5
X^2>36
Sqrtx^2>sqrt36
X>+/-6

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3(x-6)^4+11=15 solve for x

Tamiku [17]

Answer:

x = 7.07 or 4.925

Step-by-step explanation:

3(x-6)^4+11=15

3(x-6)^4=15-11

(x-6)^4=4/3

x-6 = + $\sqrt[4]{4/3}$ or - $\sqrt[4]{4/3}$

x= + $\sqrt[4]{4/3}$ +6 or - $\sqrt[4]{4/3}$ +6

x = 7.07 or 4.925

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2 years ago

Problem Page Jessica's boat has a top speed of 6 miles per hour in still water. While traveling on a river at top speed, she wen

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3 years ago

Shelby spends 2/7 of her salary on her monthly rent and has $3455 left over. How much does she spend on her rent each month?

Zepler [3.9K]

Answer:

$1382

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3 years ago

How do you reduce 36/100 using prime factorization?

Nutka1998 [239]

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3 years ago

Keenan will run 2.5 miles from his house to Jared’s house. He plans to hang out for 45 minutes before walking home. If he can ru

jenyasd209 [6]

Answer:

Kennan will be from home approximately an hour and 48 minutes.

Step-by-step explanation:

We must know that total time ( $t_{T}$ ) that Keenan will be from home is the sum of run ( $t_{R}$ ), hang out ( $t_{H}$ ) and walk times ( $t_{W}$ ), measured in hours:

$t_{T} = t_{R}+t_{H}+t_{W}$

If Keenan runs and walks at constant speed, then equation above can be expanded:

$t_{T} = \frac{x_{R}}{v_{R}}+t_{H}+ \frac{x_{W}}{v_{W}}$

Where:

$x_{R}$ , $x_{W}$ - Run and walk distances, measured in miles.

$v_{R}$ , $v_{W}$ - Run and walk speeds, measured in miles per hour.

Given that $x_{R}=x_{W} = 2.5\,mi$ , $v_{R} = 6\,\frac{mi}{h}$ , $v_{W} = 4\,\frac{mi}{h}$ and $t_{H} = 0.75\,h$ , the total time is:

$t_{T} = \frac{2.5\,mi}{6\,\frac{mi}{h} } + 0.75\,h+\frac{2.5\,mi}{4\,\frac{mi}{h} }$

$t_{T} = 1.792\,h$ ( $1\,h\,48\,m$ )

Kennan will be from home approximately an hour and 48 minutes.

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3 years ago