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ANEK [815]
2 years ago
9

33 cm 15 cm 15 cm Total surface area

Mathematics
1 answer:
LiRa [457]2 years ago
4 0

Answer:

cuboid \: total \: surface \: area \: 2(lb  + bh + lh)  l = 33 \: b = 15 \: h = 15  \\ 2(33  \times 15 + 15 \times 15 + 33 \times 15  \\ 2430

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Find the product: 377(10, 000).
Illusion [34]

Answer:

37, 700, 000

Step-by-step explanation:

377(10,000) = 377 x 10,000

= 37, 700, 000

an easy way to see it is to just add the number of zeros from 10,000 to 377. since there are four zeros, you put four zeros behind 377.

<em><u>[it's not mathematical but it helps u see the answer better. and it only works with numbers that start with 1]</u></em>

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3 years ago
What is the perimeter of the shape?
Serhud [2]

Answer:

20 cm

Step-by-step explanation:

Assuming that the side lengths you've written are the only side lengths of the shape, meaning it would be a trapezoid, all you do is add them up together.

3 + 5 + 5 + 7 = 20

3 0
2 years ago
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Help please<br> solve for x
Katen [24]
Since A and B are parallel lines, the two marked angles are congruent.
Then, 2x + 12 = x + 25
We can solve for x:
x = 13
4 0
2 years ago
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Find P(A or B) if the events are disjointed. P(A) =6/25 , and P(B) =9/25 .
nevsk [136]
Here, "disjointed" signifiies that Event A and Event B are independent; neither outcome influences the other.  In such a case, the union of A and B is given by the sum of the two probabilities P(A) and P(B).

Here, that sum is P(A or B) = 6/25 + 9/25, or 15/25, or 0.6.


8 0
3 years ago
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The length of a rectangle is three inches more than the width. the area of the rectangle is 180 inches. find the width of the re
ch4aika [34]

We have a rectangle with length L that is 3 inches more than the width W. Then we can write this as:

L=W+3

The area of the rectangle is 180 square inches.

We have to find the width W.

As the area is equal to the product of the length and the width, we can write this equation and solve for W as:

\begin{gathered} A=180 \\ L\cdot W=180 \\ (W+3)\cdot W=180 \\ W^2+3W=180 \\ W^2+3W-180=0 \end{gathered}

We have a quadratic equation. The roots of this equation will be the mathematical solutions.

We can find the roots using the quadratic formula:

\begin{gathered} W=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ W=\frac{-3\pm\sqrt[]{3^2-4\cdot1\cdot(-180)}}{2\cdot1} \\ W=\frac{-3\pm\sqrt[]{9+720}}{2} \\ W=\frac{-3\pm\sqrt[]{729}}{2} \\ W=\frac{-3\pm27}{2} \\ W_1=\frac{-3-27}{2}=-\frac{30}{2}=-15 \\ W_2=\frac{-3+27}{2}=\frac{24}{2}=12 \end{gathered}

The solutions are W = -15 and W = 12.

The first one is not valid, as W has to be greater than 0.

Then, the solution to our problem is W = 12 in.

Answer: the width is W = 12 inches.

6 0
1 year ago
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