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MrMuchimi
3 years ago
8

Consider the rectangular prism with length 3 cm, width 4 cm, and height of 6 cm. Will doubling one of the dimensions of the rect

angular prism double the surface area of the prism? Justify your answer.
Mathematics
1 answer:
yawa3891 [41]3 years ago
6 0

Answer:

NO

Step-by-step explanation:

Surface area = 2lw+ 2 lh + 2 hw  Where h = height, w= width and l = length

We will double w

SA becomes 2l(2w) + 2lh + 2h(2w)

 SA = 4lw + 2lh + 4lw

Doubling the SA

2SA = 2(2lw+ 2 lh + 2 hw)

        = 4lw + 4lh +4 hw

2SA =? SA for 2w

4lw + 4lh +4 hw =   4lw + 2lh + 4lw

This is not the same

Doubling one dimension does not double the surface area

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soldi70 [24.7K]

Hey!


RS = 15 and RT = 18. Subtract 18 from 15 to get your answer.


18 - 15 = 3


<em>That means your answer is 3</em>


\framebox{Answer = 3}

8 0
3 years ago
Read 2 more answers
What is the fewest number of quarters, nickels, and pennies that can be fairly exchanged for 2 quarters, 6 nickels, and 9 pennie
adoni [48]

Answer:

Your answer is 3 quarters, 2 nickels, and 4 pennies

Step-by-step explanation:

2 quarters plus 6 nickels plus 9 pennies is equal to 89 cents. So if we subtract 75(3 quarters) we get 14 which is less than 25 therefore we must only use nickels and pennies from now on. If we take 14 and subtract 10(2 nickels) we get 4 which less than 5 leaving us only to use pennies, which one a piece so it must be 4 pennies. Hope this helps!

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3 years ago
A quiz consists of 3 true-or-false questions and 2 multiple choice questions. The multiple choice questions have 4 options each.
Murljashka [212]
The answer would be B.96
4 0
3 years ago
Given h(x) = 2x – 3, find h(5)
kobusy [5.1K]

Answer:

h(5) = 7

Step-by-step explanation:

Step 1: Define

h(x) = 2x - 3

h(5) is x = 5

Step 2: Substitute and Evaluate

h(5) = 2(5) - 3

h(5) = 10 - 3

h(5) = 7

3 0
3 years ago
Someone help me with this
Jet001 [13]

2a. 14,9; 2b. 15,4; 2c. 30,8; 3. 32

For 2a., you have to set it up like this: csc 48° = 20⁄x [OR sin 48° = ˣ⁄20]. Then you would have to isolate the variable by getting rid of the denominator. The cosecant function has an extra step because you will get xcsc 48° = 20. As stated, isolate the variable; this time, divide by csc 48°. This is what 20 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2b., you have to set it up like this: sec 39° = ˣ⁄12 [OR cos 39° = 12⁄x. Then you would have to isolate the variable by getting rid of the denominator. The cosine function has an extra step because you will get xcos 39° = 12. As stated, isolate the variable; this time, divide by cos 39°. This is what 12 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

For 2c., you have to set it up like this: cot 64° = 15⁄x [OR tan 64° = ˣ⁄15]. Then you would have to isolate the variable by getting rid of the denominator. The cotangent function has an extra step because you will get xcot 64° = 15. As stated, isolate the variable; this time, divide by cot 64°. This is what 15 will be divided by to get your x, whereas the other side cancels out. Then you have to round to the nearest tenth degree.

Now, for 3., it is unique, but similar concept. In this exercise, we are solving for an angle measure, so we have to use inverse trigonometric ratios. So, we set it up like this: cot⁻¹ 1⅗ = m<x [OR tan⁻¹ ⅝ = m<x]. We simply input this into our calculator and we get 32,00538321°. When rounded to the nearest degree, we get 32°.

WARNING: If you use a graphing calculator, you have to input it uniquely because most graphing calculators do not have the inverse trigonometric ratios programmed in their systems. This is how you would write this: tan⁻¹ 1⅗⁻¹. You set 1⅗ to the negative first power, ALONG WITH the inverse tangent function, because without it, your answer will be thrown off. Since Cotangent and Tangent are multiplicative inverses of each other, that is the reason why the negative first power is applied ALONG WITH the inverse tangent function.

**NOTE: 1⅗ = 8⁄5

Take into consideration:

sin <em>θ</em> = O\H

cos <em>θ</em><em> </em>=<em> </em>A\H

tan <em>θ</em><em> </em>= O\A

sec <em>θ</em><em> </em>= H\A

csc <em>θ</em><em> </em>= H\O

cot <em>θ</em><em> </em>= A\O

I hope this helps you out alot, but if you are still in need of assistance, do not hesitate to let me know and subscribe to my channel [username: MATHEMATICS WIZARD], and as always, I am joyous to assist anyone at any time.

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