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Dafna11 [192]
2 years ago
12

-3/2+2 hi jjjjjjjjjjjjjjj

Mathematics
2 answers:
Lelechka [254]2 years ago
5 0
Hi and hopes this helps:

Answer: 1/2

densk [106]2 years ago
5 0

Answer:

1/2 is the answer

Step-by-step explanation:

1) Make the denominators the same.

-  \frac{3}{2}  + 2 \times  \frac{2}{2}

2) Simplify. Denominators are now the same.

-  \frac{3}{2}  +  \frac{4}{2}

3) Join the denominators.

\frac{ - 3 + 4}{2}

4) Simplify.

\frac{1}{2}

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63756/70=910.8 ft per minute

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Which expression can be used to determine the length of segment AB? On a coordinate plane, triangle A B C has points (4, 3), (ne
JulijaS [17]

Answer:

StartRoot 2 squared + 6 squared EndRoot

Step-by-step explanation:

we have

A(4,3) and B(-2,1)

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

substitute the given values

AB=\sqrt{(1-3)^{2}+(-2-4)^{2}}

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Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,
Sholpan [36]
A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
d^2=3^2+2^2
d^2=9+4
d^2=13
d= \sqrt{13}
We can conclude that the distance between my house and the movie theater is \sqrt{13}

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
d^2=6^2+5^2
d^2=36+25
d^2=61
d= \sqrt{61}
We can conclude that the distance between the police station and the airport is \sqrt{61}

C. Distance form the park to the Fire Station:
d^2=3^2+1^2
d^2=9+1
d^2=10
d= \sqrt{10}
We can conclude that the distance from the park to the Fire Station is \sqrt{10}

Distance from the stadium to the mall:
d^2=2^2+1^2
d^2=4+1
d^2=5
d= \sqrt{5}
We can conclude that the distance from the stadium to the mall is \sqrt{5}

D. What is the distance between your house and the mall?
Answer:
d^2=4^2+4^2
d^2=16+16
d^2=32
d= \sqrt{32}
d=4 \sqrt{2}
We can conclude that the distance between my house and the mall is 4 \sqrt{2}

What is the distance between the movie theater and the school?
answer:
d^2=3^2+1^2
d^2=9+1
d^2=10
d= \sqrt{10}
We can conclude that the distance between the movie theater and the school is \sqrt{10}

7 0
2 years ago
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