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3 years ago

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Find the volume of a pyramid with a square base, where the side length of the base is 6.6\text{ in}6.6 in and the height of the

pyramid is 7.7\text{ in}7.7 in. Round your answer to the nearest tenth of a cubic inch.

1 answer:

Sergeu [11.5K]3 years ago

7 0

Answer:

4.4 in^3

Step-by-step explanation:

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The student population of a school consists of 300 girls and 500 boys

denis-greek [22]

Answer:

In all you have 800 students.

The ratio of girls to boys is 300:500

500 x 300 = 150000

Step-by-step explanation:

Hope this helps?

8 0

2 years ago

Gcf of 12x^5y and #2x^6y

Murrr4er [49]

Greatest common factor is 1 for the first one and the second one is also 1

8 0

2 years ago

Mr. Alvarez makes a walkway out of 3 cement slabs. He uses 14 cubie feet of cement to make the walkway. Each square slab has a v

AysviL [449]

Answer:

Step-by-step explanation:

answer <u><em>volume cubic centimetre priority</em></u>

<em>volume=4+4=8</em>

<em>14-8=</em>

<u><em>6 cubic feet. </em></u>

<u><em>thanx</em></u>

<u><em>mark as brainlest</em></u>

6 0

2 years ago

Read 2 more answers

The sum of 3 odd consecutive integers are 93. find the 3 integers

Solnce55 [7]

Let, Your Integers = x, x+2, x+4
Now, x + x+2 + x+4 = 93
3x + 6 = 93
3x = 93 - 6
x = 87/3
x = 29
Then, x+2 = 31, & x+4 = 33

In short, Your Integers would be 29, 31, 33

Hope this helps!

4 0

2 years ago

A pan of warm water (46∘ Celsius) was put in a refrigerator. After 15 minutes, the water's temperature was 27∘ C; 15 minutes aft

Cloud [144]

Answer:

The refrigerator was at around 13 ∘C

Step-by-step explanation:

Newton's Law of Cooling:

The rate of change of a body temperature (amount of heat loss/time of loss) is directly proportional to the difference between its own temperature and the surroundings.

$\frac{T(t2)-T(t1)}{t2-t1} = -h (T(t2) - Tenv)$

T ⇒ temperature

t ⇒ time

Tenv ⇒refrigerator temperature

$\frac{T(t2)-T(t1)}{t2-t1}$ ⇒ rate of change of he temperature

-h ⇒ constant of proportionality (negative because the temperature is decreasing inside the refrigerator)

We have 3 points:

time (minutes) - Temperature (∘ C)

0 (when the pan was put in the refrigerator) - 46

15 (after 15 minutes) - 27

30 (15 minutes after the first 15 minutes) - 19

$\frac{27 - 46}{15 - 0}$ = -h (27 - Tenv)

$\frac{19 - 27}{15 - 0}$ = -h (19 - Tenv)

Now we have a system of two equations and two variables

$\frac{-19}{15} = -h (27 - Tenv)\\ \frac{19}{15} = h (27 - Tenv)\\ \frac{19}{15(27 - Tenv)} = h$

$\frac{19 - 27}{15 - 0} = -h (19 - Tenv)\\\frac{-8}{15} = -h (19 - Tenv)\\\frac{8}{15} = h (19 - Tenv)\\\frac{8}{15} = \frac{19}{15(27 - Tenv)} (19 - Tenv)\\8(27 - Tenv) = 19 (19 - Tenv)\\216 - 8 Tenv = 361 - 19 Tenv\\19 Tenv - 8 Tenv = 361 - 216\\11 Tenv = 145 \\Tenv = 145/ 11\\Tenv = 13. 18$

The refrigerator was at around 13 ∘C

6 0

2 years ago