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2 years ago

Using the Addition Method, solve for y in the following system of linear equations: 4x – 2y = 6 2x + y = 7

Mathematics

1 answer:

Lady_Fox [76]2 years ago

4 0

Answer:

y = 2

Step-by-step explanation:

Addition Method:

4x - 2y = 6

2x + y = 7

-2(2x + y) = -2(7)

4x - 2y = 6

-4x - 2y = -14

-4y = -8

y = 2

Substitution Method:

4x - 2y = 6

2x + y = 7

-2y = 6 - 4x

2y = 4x - 6

y = 2x - 3

2x + (2x - 3) = 7

4x - 3 = 7

4x = 10

x = 2.5

2(2.5) + y = 7

5 + y = 7

y = 2

Brainliest, please :)

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Read 2 more answers

The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9

mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

The height $h(t)$ of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation $h(t) = 3 +90\cdot t -16\cdot t^{2}$ , where $t$ is time in seconds. After how many seconds will the ball be 84 feet above the ground.

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

$3+90\cdot t -16\cdot t^{2} = 84$

$16\cdot t^{2}-90\cdot t +81 = 0$ (1)

By Quadratic Formula:

$t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}$

$t_{1} = 4.5\,s$ , $t_{2} = 1.125\,s$

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

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