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vazorg [7]
3 years ago
10

5) Twelve divided by a number. 6) x +10 7) 3x-5 8) 6x + 2

Mathematics
1 answer:
Maru [420]3 years ago
7 0

Answer:

5)12/x

6) a number increased by 10

7) three times a number decreased by 5

8) 6 times a number increased by 2

Step-by-step explanation: "a number" refers to a variable.

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can you help me evaluate this: 7m + 14m - 6n - 5n + 2m. why do you add 6n with 5n instead of subtract?
Mademuasel [1]

Step-by-step explanation:

The reason is because it is not just 6 it is - 6 so when there are two negative numbers they don't subtract they add if it was 6n then it was to be subtracted

21m - 11n + 2m

23m - 11n

6 0
3 years ago
Mr. Richards purchased 8 T-shirts for the volleyball team. The total cost of the T-shirts was $70.56. How much did each shirt co
bezimeni [28]

Answer:

$8.82

Step-by-step explanation:

$70.56÷8 =$8.82 total

7 0
3 years ago
X+12=31 solve for x i’m in a breakout room i have limited time to finish this
Nitella [24]

Answer: x=19

Step-by-step explanation:

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2 years ago
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miv72 [106K]

Answer:

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7 0
2 years ago
Points M, N, and P are respectively the midpoints of sides AC , BC , and AB of △ABC. Prove that the area of △MNP is on fourth of
Hunter-Best [27]

Answer:

The area of △MNP is one fourth of the area of △ABC.

Step-by-step explanation:

It is given that the points M, N, and P are the midpoints of sides AC, BC and AB respectively. It means AC, BC and AB are median of the triangle ABC.

Median divides the area of a triangle in two equal parts.

Since the points M, N, and P are the midpoints of sides AC, BC and AB respectively, therefore MN, NP and MP are midsegments of the triangle.

Midsegments are the line segment which are connecting the midpoints of tro sides and parallel to third side. According to midpoint theorem the length of midsegment is half of length of third side.

Since MN, NP and MP are midsegments of the triangle, therefore the length of these sides are half of AB, AC and BC respectively. In triangle ABC and MNP corresponding side are proportional.

\triangle ABC \sim \triangle NMP

MP\parallel BC

MP=\frac{BC}{2}

By the property of similar triangles,

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{PM^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{(\frac{BC}{2})^2}{BC^2}

\frac{\text{Area of }\triangle MNP}{\text{Area of }\triangle ABC}=\frac{1}{4}

Hence proved.

5 0
3 years ago
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