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uranmaximum [27]
3 years ago
14

Rearranging Equations: Solve the equation for y: 8x + y = 14 Solve the equation for x: -9x + 3y = 18

Mathematics
1 answer:
podryga [215]3 years ago
7 0

Answer:

y=14-8x

x= -2 + 3y/9x

Step-by-step explanation:

8x+y=14

isolate y so you get:

y=14-8x

-9x+3y=18

isolate x so you get:

-9x=18-3y

divide on both sides to get x by itself

-9x/-9x  18-3y/-9x

x= -2 + 3y/9x

Hope this helps :)

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. If ON = 4x – 6, LM = 3x + 5, NM = x – 4, and OL = 4y – 3, find the values of x and y for which LMNO must be a parallelogram.
Verizon [17]
In a parallelogram, opposite sides are congruent.

This means that
LM = ON
NM = OL

Using the equations, find x and y:
4x - 6 = 3x + 5
4x = 3x + 11
1x = 11
x = 11

x - 4 = 4y - 3
x - 1 = 4y
-1 = 3y
-1/3 = y

The values are:
x = 11
y = -1/3
5 0
3 years ago
A scatter plot was constructed and a line of best fit was drawn, 25x 80. What is the equation of this best line of fit?
Tcecarenko [31]

Answer:

y = 5x + 25

Explaination:

To get the equation of a line, first get the slope of the line.


Slope = (change in y)/(change in x)


          = (60-30)/(7-1)


          = 30/6


            = 5


Now use one point in the line (4,45) and a general point (x,y).


Slope = (change in y)/(change in x)


    5 = (y - 45)/(x - 4)


    5(x - 4) = (y - 45)


  5x - 20 = y - 45


y = 5x -20 + 45





7 0
3 years ago
Write an expression for "the difference of 10 and <br> x.
navik [9.2K]
10-x=d
Im pretty sure !!
7 0
2 years ago
Find the area of the shaded part in the given diagram
enyata [817]

Answer:

30cm^2

Step-by-step explanation:

area of large rectangle = length x width

= 5 X 9 = 45 cm^2

area of small rectangle = length x width

= 3 X 5 = 15 cm ^2

45 - 15 = 30cm^2

5 0
2 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
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