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VashaNatasha [74]
3 years ago
11

The volume of a cubic shaped water tank is 42875 m 3 , then the height of the water tank is ________m

Mathematics
2 answers:
rjkz [21]3 years ago
7 0

Answer:

35m

Step-by-step explanation:

∛42875 = 35

--------------------check

35*35*35=42875

Andrei [34K]3 years ago
3 0

Answer:

The height of this cube-shaped water tank is 35 m.

Step-by-step explanation:

The formula for the volume of a cube of side length s is V = s^3.

Here V = s^3 = 42,875 m^3.

Taking the cube root of both sides, we obtain the side length, s:

s = ∛(42,875 m^3) = 35 m

The height of this cube-shaped water tank is 35 m.

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1. Point P is between points F and G. The distance between points F and P
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Answer:

Point P is in between points F and G.

Now, here we do not have the actual locations of F and G, so i will answer this in a general way.

Suppose that F is located at 0 and G is located at some value A.

Then the distance FG = A - 0 = A.

Now, the distance between F and P is:

Distance  = (1/4)*FG = A/4

Then the location of P is equal to the location of F plus A/4

P = 0 + A/4

Now, writing this in a more general way:

P = F + FG/4

8 0
3 years ago
The sum of the measures of two complementary angles is 90°. If one angle measures 24° more than twice the measure of the​ other,
geniusboy [140]
X + y = 90

Let's make 'x' the bigger angle and 'y' the smaller.
X is 24 more than twice y.
x = 24 + 2y

Now plug that into the x of the first equation.
(24 + 2y) + y = 90

Combine like terms.
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3 0
3 years ago
For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10
professor190 [17]

The true statements are:

  • Tyson's expression is not equivalent to the original expression
  • The equivalent expression is:\frac{1}{2}g- \frac 1{5}

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}

Collect like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10}  -\frac{1}{10}

Evaluate the like terms

\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}

Tyler's equivalent expression is given as:

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}

Collect like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}

Evaluate the like terms

\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

3 0
2 years ago
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