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3 years ago

What is the simplified form of radical 50?

Mathematics

2 answers:

Mnenie [13.5K]3 years ago

6 0

Answer:

Your ultimate solution should be 5 times the square root of 2 is the simplified version of the square root of 50, which, when simplified, should be 7.05. Keep in mind that these steps can apply to any square root calculation.

Step-by-step explanation:

riadik2000 [5.3K]3 years ago

3 0

Answer:

$5\sqrt{2}$

Step-by-step explanation:

Hello!

To find the simplified form, all we have to do is factor out 50.

50: 1, 2, 5, 25, 50

Using a factor tree will also go to show that $2\cdot \:5^2$ is the prime factorization of 50.

Since it has 5², we can put it outside the radical.

Thus, the simplified form of $\sqrt{50}$ is $\boxed{5\sqrt{2}}$ .

Hope this helps!

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3 years ago

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10. 9. Solve the equation (x-2)(x+2) = 0 (4F) A. 02,0 B. O-2, 2 C.O-2 D. 2

masha68 [24]

Answer:

x=2 or x=-2

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2 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

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3 years ago

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Answer:

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