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GREYUIT [131]
3 years ago
8

To find the surface area of the pyramid, in square inches, Vikram wrote (33.2) (34.2) + 4 (one-half (34.2) (28.4)). What error d

id Vikram make? He used the same expression for the area of all four lateral faces. He used the wrong expression to represent the area of the base of the pyramid. He used the wrong value as the height when he found the area of the lateral faces. He used an expression for surface area that only finds the total area of three faces.
Mathematics
1 answer:
madam [21]3 years ago
3 0

Answer:

B. He used the wrong expression to represent the area of the base of the pyramid.

Step-by-step explanation:

Given

See attachment for pyramid

Area = (33.2)(34.2) + 4 * \frac{1}{2} * 34.4 * 28.4

Required

Vikram's error

The surface area of a square pyramid is:

Area = Base\ Area + 4 * Area \triangle

Where

Base\ Area = Length * Length

Base\ Area = 34.2 * 34.2

4 * Area \triangle = 4 * \frac{1}{2} * Base * Height

4 * Area \triangle = 4 * \frac{1}{2} * 34.2 * 28.4

So:

Area = Base\ Area + 4 * Area \triangle

Area = 34.2 * 34.2 + 4 * \frac{1}{2} * 34.2 * 28.4

By comparing the calculated expression with

Area = (33.2)(34.2) + 4 * \frac{1}{2} * 34.4 * 28.4

Option (b) is correct

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Express this to single logarithm
zzz [600]

Answer:  \log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right)

We have something in the form log(x/y) where x = q^2*sqrt(m) and y = n^3. The log is base 2.

===========================================================

Explanation:

It seems strange how the first two logs you wrote are base 2, but the third one is not. I'll assume that you meant to say it's also base 2. Because base 2 is fundamental to computing, logs of this nature are often referred to as binary logarithms.

I'm going to use these three log rules, which apply to any base.

  1. log(A) + log(B) = log(A*B)
  2. log(A) - log(B) = log(A/B)
  3. B*log(A) = log(A^B)

From there, we can then say the following:

\frac{1}{2}\log_{2}\left(m\right)-3\log_{2}\left(n\right)+2\log_{2}\left(q\right)\\\\\log_{2}\left(m^{1/2}\right)-\log_{2}\left(n^3\right)+\log_{2}\left(q^2\right) \ \text{ .... use log rule 3}\\\\\log_{2}\left(\sqrt{m}\right)+\log_{2}\left(q^2\right)-\log_{2}\left(n^3\right)\\\\\log_{2}\left(\sqrt{m}*q^2\right)-\log_{2}\left(n^3\right) \ \text{ .... use log rule 1}\\\\\log_{2}\left(\frac{q^2\sqrt{m}}{n^3}\right) \ \text{ .... use log rule 2}

8 0
2 years ago
A. 2x+5=23<br> B. x+2+5=23<br> C. x+x+5=23<br> D. 5x+2=23
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Answer:

b

Step-by-step explanation:

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3 years ago
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3 (x-5) =5<br> what is x?
azamat

Answer: 20/3

So when solving this equation first distribute our 3,we will now have   3x-15=5.

Next we will Add 15 to both sides of our equation. Now we have 3x=20.Then divide our terms,and you will have 3x/3 = 20/3

<u>This will leave you with  1x=20/3. </u>

So the X is 20/3.

      Hope that helped!

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