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Leto [7]
3 years ago
11

NEED HELP ASAP!!!! Will give brainliest!!!

Mathematics
1 answer:
iris [78.8K]3 years ago
3 0

224 cubes. Are needed 224 cubes with side lenghts of 1/4 cm to fill the prism.

The key to solve this problem is calculate the volume of both the cube and the prism.

Vcube = l³

Vprism = l*w*h

Where  l is lenght, w is width, and h is height.

Vcube = (1/4 cm)³ = 1/64 cm³

Vprism = (2 cm)(7/4 cm)(1 cm) = 7/2 cm³

Calculating how many cubes does it takes to fill the prism:

Cubes = Vprism/Vcube = (7/2 cm³)/(1/64 cm³) = 448/2 = 224

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Is (5,9) a solution to y=3x-6? Explain or show how you know.
Lisa [10]

Answer:

Yes

Step-by-step explanation:

Plug in the coordinate (5,9) into the equation

9=3(5)-6

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3 0
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Read 2 more answers
Can you please help me I promise I will mark you as the brainest I want the whole sheet please thank you
den301095 [7]

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6 0
4 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
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