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8_murik_8 [283]
3 years ago
6

What is the length of a side of a cube with a volume of 729 m3?

Mathematics
1 answer:
Law Incorporation [45]3 years ago
3 0
So, You Need To Find The Length Of One Of The Sides. So, Lets Find It Using The Cubed Method. 10*10*10 = 1000, So It Can't Be 10. 5*5*5 = 125, So It Needs To Be Bigger. 7*7*7 = 343, So Bigger. 8*8*8 = 512, So Bigger. 9*9*9 = 729m^3, So, The Side Length Of The Cube Is 9 Meters. I Hope This Helps!

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What is 6x to the third power?
murzikaleks [220]

Answer:

216x

Step-by-step explanation:

  1. Write it out: 6x^{3}  
  2. 6^{3} = 216  
  3. Add in x: 216x

I hope this helps!

8 0
3 years ago
Read 2 more answers
Help please! :(
Mekhanik [1.2K]

Answer:

5.51 meters

Step-by-step explanation:

3 0
3 years ago
A farmer wants to fence in a rectangular field that encloses 3600 square feet. One side of the field is along a river and does n
andriy [413]

Answer:

C(x) = \$3.50(\frac{x^2+7200}{x})

Step-by-step explanation:

Data provided in the question:

Area of the field = 3600 square feet

Fencing charges = $3.50 per foot

Let the side along the river be 'x' feet and the other side of the 'B'

now,

Area of rectangle = Bx = 3600 square feet

or

B = \frac{3600}{x} feet

and total length to be fenced = x + 2B

therefore,

Total cost of fencing = Fencing charges × total length to be fenced

or

Total cost of fencing = $3.50 × ( x + 2B )

substituting the value of B from (1)

Total cost of fencing, C(x) = \$3.50(x+2\times\frac{3600}{x})

or

C(x) = \$3.50(\frac{x^2+7200}{x})

4 0
3 years ago
Add an intersection the red light times normally distributed by the mean of three minutes and a standard deviation of .25 minute
Soloha48 [4]

95% of red lights last between 2.5 and 3.5 minutes.

<u>Step-by-step explanation:</u>

In this case,

  • The mean M is 3 and
  • The standard deviation SD is given as 0.25

Assume the bell shaped graph of normal distribution,

The center of the graph is mean which is 3 minutes.

We move one space to the right side of mean ⇒ M + SD

⇒ 3+0.25 = 3.25 minutes.

Again we move one more space to the right of mean ⇒ M + 2SD

⇒ 3 + (0.25×2) = 3.5 minutes.

Similarly,

Move one space to the left side of mean ⇒ M - SD

⇒ 3-0.25 = 2.75 minutes.

Again we move one more space to the left of mean ⇒ M - 2SD

⇒ 3 - (0.25×2) =2.5 minutes.

The questions asks to approximately what percent of red lights last between 2.5 and 3.5 minutes.

Notice 2.5 and 3.5 fall within 2 standard deviations, and that 95% of the data is within 2 standard deviations. (Refer to bell-shaped graph)

Therefore, the percent of  red lights that last between 2.5 and 3.5 minutes is 95%

8 0
3 years ago
At track practice,Sheila worked on the long jump. Her first jump measured 3 yards, 2 feet, 8 inches. Her second jump measured 2
jolli1 [7]
The length is 19.5 feet. this is because when you add the yards, it equals 15 feet. Next, add 1 + 2 feet from the first and second jump. 10+8 inches equal 1.5 feet. Next, add 15 feet+3 feet+1.5 feet to get 19.5 feet.
7 0
3 years ago
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