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3 years ago

9

Whoever gets this right gets 25 points and Brainlyess answer!! :)

1 answer:

dedylja [7]3 years ago

3 0

(9*4*1.5)/(1/2)^3

54/(1/8)

54*8

432

So it will take 432 cubes to fill the prism.

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When the sun is at a certain angle in the sky, a 100 foot building will cast a 25 foot shadow. How y’all is a person if he casts

uysha [10]

Answer:

6 ft

Step-by-step explanation:

The building height of 100 ft is 4 times the shadow length of 25 ft. At the same ratio, the person's height is 4 times the 1.5 ft shadow length, so is ...

4 × (1.5 ft) = 6.0 ft

The person is 6 ft tall.

4 0

3 years ago

Write any two decimals with two decimals places between 3.1 and 3.2

mart [117]

Since it is in between 3.1 and 3.2, and you are saying that it has 2 decimal places, then it could be the following numbers (choose any 2 of these you want): 3.11; 3.12; 3.13; 3.14; 3.15; 3.16; 3.17; 3.18; 3.19.

4 0

3 years ago

Read 2 more answers

The recommended angle for a wheelchair ramp is 5 degrees. If the rise of the ramp to go up the steps is 5 feet, find the horizon

dybincka [34]

Answer:

$57.2\ ft$

Step-by-step explanation:

we know that

The formula of slope is "rise over run", where the "rise" (means change in y, up or down) and the "run" (means change in x, left or right)

In this problem

The tangent of angle of 5 degrees is equal to the quotient of "rise over run"

Let

y ----> the rise of the ramp

x ----> the run of the ramp

$tan(5^o)=\frac{y}{x}$

we have

$y=5\ ft$

substitute and solve for x

$tan(5^o)=\frac{5}{x}$

$x=\frac{5}{tan(5^o)}$

$x=57.2\ ft$

5 0

4 years ago

In the parallelogram below,

Assoli18 [71]

Step-by-step explanation:

5X _17 = 3 X +7

2X = 24

X = 12

5 0

3 years ago

Please Help Quickly!!!<br><br> Find the limit if f(x) = x^3

jeka94

Answer:

Option b. 12

Step-by-step explanation:

This exercise asks us to find the derivative of a function using the definition of a derivative.

Our function is $f(x) = x^{3}$ . Therefore:

$f(2+h) = (2+h)^{3}$

$f(2) = (2)^{3} = 8$

Then:

$\lim_{h \to \0} \frac{f(2+h)-f(2)}{h}=\lim_{h \to \0} \frac{(2+h)^{3}-8}{h}$

Expanding:

$\lim_{h \to \0} \frac{(2+h)^{3}-8}{h} =\lim_{h \to \0} \frac{8+ h^{3} +6h(2+h) -8}{h} =\lim_{h \to \0} \frac{h^{3} +6h(2+h)}{h}$

$\lim_{h \to \0} \frac{h^{3}+ 6h(2+h)}{h} =\lim_{h \to \0} h^{2} + 6(2+h)$

Now, if x=0:

$\lim_{h \to \0} \frac{f(2+h)-f(2)}{h} = (0)^{2} +6(2+0) = 12$

4 0

3 years ago