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2 years ago

It says find the volume of the prism.

Mathematics

1 answer:

Misha Larkins [42]2 years ago

7 0

Answer:

the answer is 3 1/8 cubic cm

Step-by-step explanation:

brainliest if correct hope this helps have a blessed day! :)

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Solve the inequality 42<-6d

pentagon [3]

Answer:

d < - 7

Step-by-step explanation:

given

42 < - 6d ( divide both sides by - 6 )

Remembering to reverse the inequality symbol as a consequence

- 7 > d ⇒ d < - 7

6 0

3 years ago

Larry has a new mortgage of $150,000 with an interest rate of 10% per year. What is the balance after 3 monthly payments of $1,3

GaryK [48]

Https://answers.yahoo.com/question/index?qid=20120624121412AAV4Cjw
Hope that helps!
If the link doesn't work if you click on it, then just copy and paste.

3 0

2 years ago

a cake divided into 10 equal slices.kim ate 3/5 of 1/2 of cake. what part of the whole cake did kim ate?

joja [24]

Answer:

3/10 of the whole cake

Step-by-step explanation:

When you divide a cake in halve you get 5 pieces on each side, if Kim ate 3/5 of 1/2 of the cake, then she ate 3 pieces, so she ate 3/10 of the cake, and other way to figure this out is multiplying the two fractions, 3/5 x 1/2 = 3/10

8 0

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

8 0

3 years ago

What is the height of this triangular prism?

chubhunter [2.5K]

The height will be four, becaus 5 is the height of the base not the whole prism. You can tell be beacause when you find the area of the base and multiply it by the height you multiply it by 4 not 5.

5 0

3 years ago