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irga5000 [103]
3 years ago
11

if one face of a cube has an area of 6.75 square inches , what is the surface area of the entire cube ?​

Mathematics
1 answer:
Travka [436]3 years ago
3 0

Answer:

40.5

Step-by-step explanation:

a cube has 6 faces so 6.75 x 6 = 40.5

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D varies as R and S, and inversely as t. D=12, R=3, S=20, and t=5, find D when R=15, S=4 and t=8.
zavuch27 [327]

Given:

D varies as R and S, and inversely as t.

D=12, R=3, S=20, and t=5

To find:

The value of D when R=15, S=4 and t=8.

Solution:

It is given that D varies as R and S, and inversely as t. So,

D\propto \dfrac{RS}{t}

D=\dfrac{kRS}{t}                ...(i)

Where, k is the constant of proportionality.

We have, D=12, R=3, S=20, and t=5. Substituting these values in (i), we get

12=\dfrac{k(3)(20)}{5}

12=12k

\dfrac{12}{12}=k

1=k

Substituting k=1 in (i), we get the required equation.

D=\dfrac{(1)RS}{t}

D=\dfrac{RS}{t}

We need to find D when R=15, S=4 and t=8. Substituting R=15, S=4 and t=8 in the above equation, we get

D=\dfrac{(15)(4)}{8}

D=\dfrac{60}{8}

D=7.5

Therefore, the required value of D is 7.5.

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Answer:

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Step-by-step explanation:

(-7 , 8)

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A student concluded that the inequality −3+2y≤4x is equivalent to the inequality y≥2x+32, as shown below. Describe and correct t
timofeeve [1]

Answer:

-3 +2y \leq 4x

2y \leq 4x+3

y \leq 2x + \frac{3}{2}

And that's different from the claim of the student that:

y \geq 2x +\frac{3}{2}

The error of the student is that he/she changes the sign of the inequality from \leq to \geq and that's not possible since we don't multiply both sides of the equation by -1

Step-by-step explanation:

For this case we have the following inequality:

-3 +2y \leq 4x

We want to rewrite the last expression with y in the left and x in the right so we can begin adding 3 in both sides of the inequality and we got:

2y \leq 4x+3

Now we can divide both sides of the inequality by 2 and we got:

y \leq 2x + \frac{3}{2}

And that's different from the claim of the student that:

y \geq 2x +\frac{3}{2}

The error of the student is that he/she changes the sign of the inequality from \leq to \geq and that's not possible since we don't multiply both sides of the equation by -1

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