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Soloha48 [4]
4 years ago
9

3x-2y=14 2x-3y=11 solve algebraically

Mathematics
1 answer:
VARVARA [1.3K]4 years ago
3 0

Answer: x = 4, y = - 1

Step-by-step explanation:

The given system of simultaneous equations is expressed as

3x - 2y = 14 - - - - - - - - - - - - 1

2x - 3y = 11- - - - - - - - - - - - - 2

The first step is to decide on which variable to eliminate. Let us eliminate x. Then we would multiply both rows by numbers which would make the coefficients of x to be equal in both rows.

Multiplying equation 1 by 2 and equation 2 by 3, it becomes

6x - 4y = 28

6x - 9y = 33

Subtracting, it becomes

5y = - 5

y = - 5/- 5 = - 1

The next step is to substitute y = - 1 into any of the equations to determine x.

Substituting y = - 1 into equation 2, it becomes

2x - 3 × - 1 = 11

2x + 3 = 11

2x = 11 - 3 = 8

x = 8/2

x = 4

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What is 56÷34×y and add 12
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Hey there! :)

PARENTHESES
EXPONENTS
MULTIPLICATION
DIVISION
ADDITION
SUBTRACTION

First is dividing

56 ÷ 34 = 56/34

56/34 = 28/17

56 ÷ 2 = 28
34 ÷ 2 = 17

Then multiply by y

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7 0
3 years ago
1.
Margarita [4]

Answer:

The ball throw into the air with an initial velocity of 22 meters per second is 27.4 meters above the ground after 3 seconds.

Step-by-step explanation:

The quadratic function h(t) = -4.9t^2 + 22t + 5.5 represents the height of the ball above the ground, h(t), in meters, with respect to time, t, in seconds.

To find h(3), substitute t = 3 into the function expression:

h(3)=-4.9\cdot 3^2+22\cdot 3+5.5\\ \\=-4.9\cdot 9+66+5.5\\ \\=-44.1+71.5\\ \\=27.4

Meaning: the ball throw into the air with an initial velocity of 22 meters per second is 27.4 meters above the ground after 3 seconds.

5 0
4 years ago
Given the position function s(t), s(t) = t2 + 3t + 1, where s is measured in meters and t is in seconds, find the velocity and a
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Answer:

  11 m/s; 2 m/sec²

Step-by-step explanation:

v(t) = s'(t) = 2t +3

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a(t) = v'(t) = 2 . . . . m/s²

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Velocity and acceleration are 11 m/s and 2 m/s².

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