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Karo-lina-s [1.5K]
2 years ago
12

What is the volume of the following rectangular prism?

Mathematics
2 answers:
inessss [21]2 years ago
5 0

Answer:

Length * Height * Width, or V = L * H * W. Ex: V = 5 in.

Nina [5.8K]2 years ago
4 0

Answer:

28

Step-by-step explanation:

We are given the area of a base and the width of the prism.

Volume = area of base * width

Volume = 8/3 * 21/2 = 28

You might be interested in
The circumference of the circle is about ? Inches
stiv31 [10]

Answer:

18.84 inches

Step-by-step explanation:

The circumference is given by

C = pi d  where d is the diameter

C = 3.14 *6

C = 18.84

3 0
2 years ago
6. Michael drove a small car
alekssr [168]

Answer:

unknown

Step-by-step explanation:

15 times 3.14

8 0
3 years ago
Read 2 more answers
Evaluate the limit
wel

We are given with a limit and we need to find it's value so let's start !!!!

{\quad \qquad \blacktriangleright \blacktriangleright \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

But , before starting , let's recall an identity which is the <em>main key</em> to answer this question

  • {\boxed{\bf{a^{2}-b^{2}=(a+b)(a-b)}}}

Consider The limit ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}}

Now as directly putting the limit will lead to <em>indeterminate form 0/0.</em> So , <em>Rationalizing</em> the <em>numerator</em> i.e multiplying both numerator and denominator by the <em>conjugate of numerator </em>

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}\times \dfrac{\sqrt{x}+\sqrt{3\sqrt{x}-2}}{\sqrt{x}+\sqrt{3\sqrt{x}-2}}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-\sqrt{3\sqrt{x}-2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}{(x^{2}-4^{2})(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Using the above algebraic identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x})^{2}-(\sqrt{3\sqrt{x}-2})^{2}}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-(3\sqrt{x}-2)}{(x-4)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}+2}{\{(\sqrt{x})^{2}-2^{2}\}(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , here we <em>need</em> to <em>eliminate (√x-2)</em> from the denominator somehow , or the limit will again be <em>indeterminate </em>,so if you think <em>carefully</em> as <em>I thought</em> after <em>seeing the question</em> i.e what if we <em>add 4 and subtract 4</em> in <em>numerator</em> ? So let's try !

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{x-3\sqrt{x}-2+4-4}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(x-4)+2+4-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , using the same above identity ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+6-3\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+3(2-\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take minus sign common in <em>numerator</em> from 2nd term , so that we can <em>take (√x-2) common</em> from both terms

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)-3(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , take<em> (√x-2) common</em> in numerator ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-2)\{(\sqrt{x}+2)-3\}}{(\sqrt{x}-2)(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Cancelling the <em>radical</em> that makes our <em>limit again and again</em> <em>indeterminate</em> ;

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{\cancel{(\sqrt{x}-2)}\{(\sqrt{x}+2)-3\}}{\cancel{(\sqrt{x}-2)}(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}+2-3)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

{:\implies \quad \displaystyle \sf \lim_{x\to 4}\dfrac{(\sqrt{x}-1)}{(\sqrt{x}+2)(x+4)(\sqrt{x}+\sqrt{3\sqrt{x}-2})}}

Now , <em>putting the limit ;</em>

{:\implies \quad \sf \dfrac{\sqrt{4}-1}{(\sqrt{4}+2)(4+4)(\sqrt{4}+\sqrt{3\sqrt{4}-2})}}

{:\implies \quad \sf \dfrac{2-1}{(2+2)(4+4)(2+\sqrt{3\times 2-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{6-2})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+\sqrt{4})}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(2+2)}}

{:\implies \quad \sf \dfrac{1}{(4)(8)(4)}}

{:\implies \quad \sf \dfrac{1}{128}}

{:\implies \quad \bf \therefore \underline{\underline{\displaystyle \bf \lim_{x\to 4}\dfrac{\sqrt{x}-\sqrt{3\sqrt{x}-2}}{x^{2}-16}=\dfrac{1}{128}}}}

3 0
2 years ago
Read 2 more answers
What is the area of a garden that measures 4mx4mx6mx4mx10mx8m
topjm [15]

The area of the garden is 120 square metre

<h2>Explanation:</h2>

The picture of the garden is shown below. In this problem, we have:

  • One square
  • Two rectangles

Recall that the area of a square is given by:

A_{s}=L^2 \\ \\ A_{s}:Area \ of \ a \ square \\ L:Side \ of \ the \ square

On the other hand, the area of a rectangle is given by:

A_{r}=L_{1}\times L_{2} \\ \\ \\ A_{r}:Area \ of \ a \ rectangle \\ \\ L_{1}:Side \ 1 \ of \ the \ rectangle \\ \\ L_{2}:Side \ 2 \ of \ the \ rectangle

Therefore:

FOR THE SQUARE:

L=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{s}=4^2 \\ \\ \boxed{A_{s}=16m^2}

FOR THE RECTANGLE 1:

L_{1}=6m \\ L_{2}=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{1}}=6(4) \\ \\ \boxed{A_{r_{1}}=24m^2}

FOR THE RECTANGLE 2:

L_{1}=10m \\ L_{2}=8m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{2}}=10(8) \\ \\ \boxed{A_{r_{2}}=80m^2}

Finally, the area of a garden is the sum of the three areas:

A_{total}=A_{s}+A_{r_{1}}+A_{r_{2}} \\ \\ A_{total}=16m^2+24m^2+80m^2 \\ \\ A_{total}=16+24+80 \\ \\ \boxed{A_{total}=120m^2}

<h2>Learn more:</h2>

Area of a pizza: brainly.com/question/12878495

#LearnWithBrainly

4 0
3 years ago
MULTIPLYING FRACTIONS YES IM AN IDOIT SO HELP ME PLEASE XOXOXOXOXOXOXOXOXOXOXOXXO
Allisa [31]
2miles=20km
20km/4=5km
5km=0,5mile
Runned 0.5 miles in total
3 0
2 years ago
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