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Illusion [34]
3 years ago
13

Find the surface area of the rectangular prism.

Mathematics
1 answer:
amm18123 years ago
5 0

Answer:

70 m²

Step-by-step explanation:

Red sides:

2 x ( 5 x 5 ) = 2 x 25 = 50

Blue side :

2 x ( 1 x 5) = 2 x 5 = 10

Green :

2 x ( 1 x 5 ) = 2 x 5 = 10

Total surface area:

50 + 10 + 10 = 70

Units : m²

Surface Area = 70 m²

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

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260 units2 is the answer

3 0
3 years ago
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Work out m and c for the line: y = x − 6
Nataly [62]

Answer:

As y = mx + c (standard form)

Thus relating the equation , we get

m = 1

c = -6

8 0
3 years ago
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Point Q is the image of Q (-5,1) under a translation by 6 units to the right and 2 units down.
Alex787 [66]

Answer:

(1,-1)

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The graph is being shifted 6 units to the right which affects the x-value and it is also being affected 2 units down which affects the y-value.

6 0
3 years ago
Which equation shows the correct use of the Power of Products Property?
Vadim26 [7]
The power of products property states that for number enclosed in a bracket or parenthesis, if it is raised to a power, it must be multiplied to the power of the enclosed number no matter how different the base is. You cannot add it because it is not raised. You can only add it if they have the same base. But in this problem, you will just multiply it. The breakdown of the solution to this problem is shown below. So,
<span><span>• (2x⁵y²)³=(21x3x5*3y2*3) = 6x15y6</span><span>

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7 0
3 years ago
In a bag of m&amp;m's there are 5 brown 6 yellow 4 blue 3 green and 2 orange. What's the probability of getting 3 yellow m&amp;m
olasank [31]
There are 5+6+4+3+2=20 m&m's in the bag.
Calculate in how many ways you can choose 3 m&m's from 20:
_{20} C _3=\frac{20!}{3!(20-3)!}=\frac{20!}{3! \times 17!}=\frac{17! \times 18 \times 19 \times 20}{6 \times 17!}=\frac{18 \times 19 \times 20}{6}=3 \times 19 \times 20= \\&#10;=1140

There are 6 yellow m&m's.
Calculate in how many ways you can choose 3 m&m's from 6:
_6 C _3 = \frac{6!}{3!(6-3)!}=\frac{6!}{3! \times 3!}=\frac{3! \times 4 \times 5 \times 6}{3! \times 6}=\frac{4 \times 5 \times 6}{6}=4 \times 5=20

The probability is the number of ways of choosing 3 m&m's from 6 m&m's divided by the number of ways of choosing 3 m&m's from 20 m&m's.
P=&#10;\frac{20}{1140}=\frac{20 \div 20}{1140 \div 20}=\frac{1}{57}

The probability is 1/57.
4 0
3 years ago
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