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alexdok [17]
3 years ago
12

To get from the post office to the library, Sindy walked due north and then due east. The distance Sindy walked east is 20 meter

s more than the distance she walked north. If Sindy were able to walk directly from the post office to the library in a straight line, she would have walked 150 meters. Suppose x represents the distance Sindy walked north. Which equation can be used to find x?
x^2+(x+20)^2=150^2
202=x^2-150^2
x+(x+20)=150
x^2+(x+150)^2=20^2
Mathematics
2 answers:
kaheart [24]3 years ago
6 0
A. x∧2+(x+20∧2=150∧2  is correct.

because you need to plug the numbers into the pathagoreum theorum which is 
a∧2+b∧2=c∧2

150 is our c because it is the side right accross from the right angle.

which number is a and which number is b does not matter.

x is the distance north so we can assign that to a and the distace east is north plus 20 so we can assign b to x+20.
masya89 [10]3 years ago
5 0

Answer:

150 is our c because it is the side right accross from the right angle.

which number is a and which number is b does not matter.

x is the distance north so we can assign that to a and the distace east is north plus 20 so we can assign b to x+20.

a. x∧2+(x+20∧2=150∧2  is correct.

because you need to plug the numbers into the pathagoreum theorum which is 

a∧2+b∧2=c∧2

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Draw a frequency polygon for the following data:
const2013 [10]

Answer:

See attachment

Step-by-step explanation:

Given

\begin{array}{ccccccc}{Marks} & {0-10} & {10-20} & {20-30} & {30-40} & {40-50}  & {50-60}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

Required

The frequency polygon

We have:

\begin{array}{ccccccc}{Marks} & {0-10} & {10-20} & {20-30} & {30-40} & {40-50}  & {50-60}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

First, we calculate the midpoint of each class

\begin{array}{ccccccc}{Midpoint} & {(0+10)/2} & {(10+20)/2} & {(20+30)/2} & {(30+40)/2} & {(40+50)/2}  & {(50+60)/2}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

\begin{array}{ccccccc}{Midpoint} & {5} & {15} & {25} & {35} & {45}  & {55}\ \\ {Students} & {7} & {15} & {22} & {30} & {16} & {10} \ \end{array}

Lastly, we plot the midpoint against the frequency of students (see attachment)

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