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2 years ago

Determine the equation

Mathematics

1 answer:

sergejj [24]2 years ago

8 0

Answer:

$y=\frac{3}{4} x-\frac{5}{4}$

Step-by-step explanation:

two points on the line are

(-1, -2)

(-5, -5)

The slope:

$m=\frac{-2-(-5)}{-1-(-5)} =\frac{-2+5}{-1+5} =\frac{3}{4}$

The equation:

with (-1,-2)

$y-(-2)=\frac{3}{4} (x-(-1))$

$y+2=\frac{3}{4} x+\frac{3}{4}$

$y=\frac{3}{4} x+\frac{3}{4} -\frac{8}{4}$

$y=\frac{3}{4} x-\frac{5}{4}$

Hope this helps

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Jasmine took a cab home from her office.the cab charged a flat fee of $4 plus $3 per mile jasmine $32 for the tip. how far does

Annette [7]

32 = 4+3x
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3 years ago

How can I find the value of X?

BARSIC [14]

Answer:

Step-by-step explanation:

Remark

The first thing you must do is notice that <D = <BEA. Mark it that way on your diagram.

The Second thing you need to notice is that <BEA and <A are equal because <A and <D are marked as equal.

So both the large triangle and the small one are isosceles. because 2 of the 3 angles are equal.

Now are you ready for this? That means that BA = 10 because BA is opposite one of the 2 equal angles in the small triangle. So now you are ready to set up a proportion.

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AB = 10

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2 years ago

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 $\sqrt{10}$