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2 years ago

What the square of 8^2

Mathematics

2 answers:

Novosadov [1.4K]2 years ago

3 0

Answer:

8^2 = 64

unless you mean 8^2's square root then it is 8

Step-by-step explanation:

olchik [2.2K]2 years ago

3 0

Answer:

8^2 = 64

Step-by-step explanation:

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You need to write 0.63 as a fraction. How do you choose the denominator?

Tamiku [17]

I would use 100 as the denominator because it can not be simplified as a whole number

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3 years ago

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Choose all the numbers that 810 is divisible by.

Aleks04 [339]

810/2= 405

810/3= 270

810/4= 202.5 (not this one)

810/5= 162

810/6= 135

810/7= 115.71 ( not this one)

810/8= 101.25 (not this one)

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810/10= 81

810/11= 73.63 (not this one)

I hope this helps you!

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3 years ago

G(n)=-4n+2; find g(8)<br>

yarga [219]

Answer:

-30

Step-by-step explanation:

In g(8) the 8 = n. So, substitute 8 in for n in the expression -4n + 2

g(8) = -4(8) + 2 = -32 + 2 = -30

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3 years ago

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The cost of 5 diesels is $1025. Calculate the cost of 17diesels

Talja [164]

To calculate the cost of 1 diesel divide the cost of 5 by 5 to get 1 or $1025/5=205
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3 years ago

A tank contains 30 lb of salt dissolved in 300 gallons of water. a brine solution is pumped into the tank at a rate of 3 gal/min

sesenic [268]

$A'(t)=(\text{flow rate in})(\text{inflow concentration})-(\text{flow rate out})(\text{outflow concentration})$
$\implies A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\left(2+\sin\dfrac t4\right)\dfrac{\text{lb}}{\text{gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$
$A'(t)+\dfrac1{100}A(t)=6+3\sin\dfrac t4$

We're given that $A(0)=30$ . Multiply both sides by the integrating factor $e^{t/100}$ , then

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$\left(e^{t/100}A(t)\right)'=6e^{t/100}+3e^{t/100}\sin\dfrac t4$
$e^{t/100}A(t)=600e^{t/100}-\dfrac{150}{313}e^{t/100}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+C$
$A(t)=600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)+Ce^{-t/100}$

Given that $A(0)=30$ , we have

$30=600-\dfrac{150}{313}\cdot25+C\implies C=-\dfrac{174660}{313}\approx-558.02$

so the amount of salt in the tank at time $t$ is

$A(t)\approx600-\dfrac{150}{313}\left(25\cos\dfrac t4-\sin\dfrac t4\right)-558.02e^{-t/100}$

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3 years ago

What the square of 8^2​

What the square of 8^2