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4 years ago

9

Find the total surface area of the cube box below

2 answers:

zaharov [31]4 years ago

5 0

The surface area of the cube is 24 feet.

dem82 [27]4 years ago

4 0

We know a cube has 6 equivalent sides. One side has an edge of 2 ft so the other edges are all 2 ft so 6(2 x 2) or 2 x 2 + 2 x 2 + 2 x 2 + 2 x 2 + 2 x 2 + 2 x 2 = 24ft.

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What’s the answer????

enot [183]

The answer should be 4

7 0

4 years ago

Read 2 more answers

How manys days makes a leap year?

dem82 [27]

Answer: In an ordinary year, if you were to count all the days in a calendar from January to December, you'd count 365 days. But approximately every four years, February has 29 days instead of 28. So, there are 366 days in the year. This is called a leap year.

Step-by-step explanation:

3 0

3 years ago

Can someone please help me

djyliett [7]

Slope = (10 - 5)/(-9 + 15) = 5/6

y = mx + b where m = slope and b = y-intercept

b = y - mx

b = 5 - (5/6) (-15)

b = 5 +12.5

b = 17.5

Equation y = 5/6 x + 17.5

Y-intercept = 17.5

X-intercept when y = 0

So

5/6 x + 17.5 = 0

5/6 x = -17.5

x = (-17.5) (6/5)

x = -21

Answer

Y intercept (0, 17.5 )

X intercept (-21 , 0)

3 0

3 years ago

Loren drove 200 miles at a certain rate, and his wife, Lois, drove 100 miles at a rate 10 mph slower. If Loren had driven for th

NeX [460]

As long as Loren drove, the law of motion was

$200 = st_1 \implies t_1 = \dfrac{200}{s}$

As long as Loid drove, the law of motion was

$100 = (s-10)t_2 \implies t_2 = \dfrac{100}{s-10}$

So, the total time they took is

$t_1+t_2=\dfrac{200}{s}+\dfrac{100}{s-10}$

Had Loren driven the whole time, the law of motion would have been

$300=st_3 \implies t_3 = \dfrac{300}{s}$

And we know that this time would have been 30 minutes (i.e. 0.5 hours) faster. So, we have

$t_3 = t_1+t_2-0.5$

This translates into

$\dfrac{300}{s}=\dfrac{200}{s}+\dfrac{100}{s-10}-\dfrac{1}{2}$

If we subtract 200/s from both sides, we have

$\dfrac{100}{s}=\dfrac{100}{s-10}-\dfrac{1}{2}$

We can simplify the right hand side by summing the two fractions:

$\dfrac{100}{s-10}-\dfrac{1}{2} = \dfrac{200-(s-10)}{2(s-10)}=\dfrac{210-s}{2(s-10)}$

So, we have to solve

$\dfrac{100}{s}=\dfrac{210-s}{2(s-10)}$

If we cross multiply the denominators, we have

$200(s-10)=s(210-s) \iff 200s-2000=210s-s^2 \iff s^2-10s-2000=0$

Which yields the solutions

$s=-40,\quad s=50$

We accept the positive solution, because the negative would mean to travel backwards, so Loren's rate was 50mph

5 0

3 years ago

Read 2 more answers

Help pls iready math equation

Svet_ta [14]

Answer:

4 and 5

Step-by-step explanation:

n + 1 > 4

n > 4 - 1

n > 3

For the set { 1, 2 , 3 , 4, 5 }, 4 and 5 are greater than 3.

Hence, 4 and 5 makes the inequality n + 1 > 4 true.

4 0

3 years ago