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Leno4ka [110]
2 years ago
6

Surface= Area= Help please thanks

Mathematics
1 answer:
BigorU [14]2 years ago
8 0

Surface area of the rectangular solid = 416 in.².

Volume = 480 in.³.

<h3>What is the Surface Area and Volume of a Rectangular Solid?</h3>

Surface area = 2(wl+hl+hw)

Volume = (length)(width)(height).

Given the following:

Length (l) = 12 in.

Width (w) = 10 in.

Height (h) = 4 in.

Surface area = 2(wl+hl+hw) = =2·(10·12+4·12+4·10) = 416 in.².

Volume = (12)(10)(4) = 480 in.³.

Learn more about the surface area of rectangular solid on:

brainly.com/question/1310421

#SPJ1

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Answer:

0.6 %

Step-by-step explanation:

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5 0
3 years ago
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Wittaler [7]

Part a.

The domain is the set of x values such that x \ge -\frac{1}{2}, basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve 2x+1 \ge 0 for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.

If you want the domain in interval notation, then it would be \Big[ -\frac{1}{2} , \infty \Big) which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.

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Part b.

I'm going to use "sqrt" as shorthand for "square root"

f(x) = sqrt(2x+1)

f(10) = sqrt(2*10+1) ... every x replaced by 10

f(10) = sqrt(20+1)

f(10) = sqrt(21)

f(10) = 4.58257569 which is approximate

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Part c.

f(x) = sqrt(2x+1)

f(x) = sqrt(2(x)+1)

f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)

f(x+2a) = sqrt(2x+4a+1) .... distribute

we can't simplify any further

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3 years ago
The box is 2.1 feet long,2.7 feet wide and 3.2 feet high. What is closest to the total surface area of the box?
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3 years ago
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Scorpion4ik [409]
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3 years ago
show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this
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Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

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• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

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Hence roots of the equation are always real for a, b, k ∈ R


           

8 0
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