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Delvig [45]
2 years ago
12

Which terms could have a greatest common factor of 5m2n2?.

Mathematics
1 answer:
ozzi2 years ago
3 0

Answer:

a common factor would be 5m^4n^3 or 15m^2n^2

Step-by-step explanation:

not sure what terms they want you to choose from but anything that starts with a number (5,10,15,) and "m" would any number 5 would go into

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If a person drives his car in the speed 50 miles per hour, how far can he cover the distance in 2.5 hours ?
ad-work [718]

Answer:

Hey there!

50 miles = 1 hour

x miles = 2.5 hours

50/1= x/2.5

x=2.5x50

x=125

The car can go 125 miles.

Hope this helps :)

4 0
3 years ago
Read 2 more answers
Helen paid $246 for three jeans from her favourite clothing store. How much did she pay for each jeans she bought?
Natali5045456 [20]

Answer:

estimate: $80

exact amount paid: $82

Step-by-step explanation:

If you round $246 to 240, then divide it by 3, you should get $80. The cost per pair should be around $80 each

For the exact answer you take the amount spent (246) and divide it by the amount of jeans she bought (3).

246/3 = 82

7 0
2 years ago
(7,-1) and (6, 6)
vova2212 [387]

Answer:

y=-7x+48 slope would be -7

Step-by-step explanation:

Used a Regression Calculator

3 0
2 years ago
Celina and Richard spend a certain amount of money from their money box each month to buy plants.
loris [4]

Answer:

100

Step-by-step explanation:

5 0
3 years ago
This hyperbola is centered at theorigin. Find its equation.Foci: (-2,0) and (2,0)Vertices: (-1,0) and (1,0)
beks73 [17]

SOLUTION

From the question, the center of the hyperbola is

\begin{gathered} (h,k),\text{ which is } \\ (0,0) \end{gathered}

a is the distance between the center to vertex, which is -1 or 1, and

c is the distance between the center to foci, which is -2 or 2.

b is given as

\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}

But equation of a hyperbola is given as

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

Substituting the values of a, b, h and k, we have

\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}

Hence the answer is

\frac{x^2}{1}-\frac{y^2}{3}=1

5 0
10 months ago
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