All categories

3 years ago

What is the surface area in centimeters of a cube with a side lengths of 1/8 centimeters

Mathematics

2 answers:

ipn [44]3 years ago

4 0

Answer:

1/8 cm = 0.125 cm = 1.25 mm

6 * (1.25 mm)^2 = 9.375 mm^2

nydimaria [60]3 years ago

4 0

Answer:

6/64cm

Step-by-step explanation:

Surface area formula: 6s^2

plug in: 6(1/8)^2

solve: 6(1/64)

solve again: 6/64

You might be interested in

Prestige Car Rentals charges $44 per day plus 6 cents per mile to rent a mid-sized vehicle. Gateway Auto charges $35 per day plu

Mama L [17]

Make two equations from the information. Then graph them and see where they intersect. The answer is 62 miles. Plug 62 back into both equations and they should both be the same number.

6 0

3 years ago

Read 2 more answers

An automobile insurance company issues a one-year policy with a deductible of 500. The probability is 0.8 that the insured autom

Effectus [21]

Answer:

y₀.₉₅ = 3659

Step-by-step explanation:

P( no accident ) = 0.8

P( one accident ) = 0

deductible = 500

mean = 3000

Determine the 95th percentile of the insurance company payout

Assuming : y =company payout , x =amount of loss incurred due to accident

Then :

P( x < 500 ) = 0.2 ( 1 - e^-500/3000)

= 0.2 ( 1 - e^-1/6 )

95th percentile =

= P( y < y₀.₉₅ ) 0.95

P( y = 0 ) = 0.8 + 0.2 ( 1 - e^-1/6 ) = 0.8307

attached below is the remainder of the solution

5 0

3 years ago

Anyone can help me with my math homework please?

Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

hello,

so we know y in terms of t and x in terms of t and we need to find y in terms of x

$x=21t^2\sqrt{x}=\sqrt{21}*t \ \ as \ \ t>=0 \ \ So\\t=\sqrt{\dfrac{x}{21}}$

and then

$y=f(x)=3\sqrt{\dfrac{x}{21}}+5=\sqrt{\dfrac{9x}{21}}+5=\sqrt{\dfrac{3x}{7}}+5$

hope this helps

4 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 $y=2$ .