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3 years ago

What is the surface area?

Mathematics

1 answer:

Blizzard [7]3 years ago

8 0

Answer:

42 square yards

Step-by-step explanation:

T.S.A of cube

$= 6l {}^{2} \\$

where l = 7yd

$= 6(7) \\ = 42 \: yd {}^{2}$

thus, the surface area of cube is 42 square yards

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12) As of August, 2011, Russia has had 21 rocket launch failures out of 743 total launches. To the nearest whole number, how man

bogdanovich [222]

Answer:

Step-by-step explanation:

It is given that, in Russia there are 21 rocket launch failures out of 743 total launches. We need to find no of failures for the next 800 launches.

21 rocket = 743 launches

1 launch = $\dfrac{21}{743}\ \text{failures}$

For 800 launches,

$=\dfrac{21}{743}\ \text{failures}\times 800\\\\=22.611\ \text{failures}$

= 23 failures

So, there are 23 failures in next 800 launches.

5 0

3 years ago

Which of the following sets could be the sides of a right triangle?

coldgirl [10]

Answer: Choice B) {3, 5, sqrt(34)}

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

Explanation:

We can only have a right triangle if and only if a^2+b^2 = c^2 is a true equation. The 'c' is the longest side, aka hypotenuse. The legs 'a' and 'b' can be in any order you want.

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For choice A,

a = 2

b = 3

c = sqrt(10)

So,

a^2+b^2 = 2^2+3^2 = 4+9 = 13

but

c^2 = (sqrt(10))^2 = 10

which is not equal to 13 from above. Cross choice A off the list.

-----------

Checking choice B

a = 3

b = 5

c = sqrt(34)

Square each equation

a^2 = 3^2 = 9

b^2 = 5^2 = 25

c^2 = (sqrt(34))^2 = 34

We can see that

a^2+b^2 = 9+25 = 34

which is exactly equal to c^2 above. This confirms the answer.

-----------

Let's check choice C

a = 5, b = 8, c = 12

a^2 = 25, b^2 = 64, c^2 = 144

So,

a^2+b^2 = c^2

25+64 = 144

89 = 144

which is a false equation allowing us to cross choice C off the list.

7 0

3 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .