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aliina [53]
3 years ago
6

Find the equation for the line that passes through the point (2,2), and that is parallel to the line with the equation y=3/4x-4

Mathematics
1 answer:
telo118 [61]3 years ago
6 0

Answer: y=3/4x  + 1/2

Step-by-step explanation: Parallel lines must have the same slope, so you already know the equation must be something like y=3/4x+b. You are given a point that the line goes thru(2,2), so you can plug this point into the equation to solve for b.

(2)=3/4(2) + b

2=6/4 + b

2= 3/2 + b

b= 2-3/2

b=4/2- 3/2

b=1/2

so y=3/4x + 1/2

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PLEASE HELP! 20 POINTS 1) A ball is thrown starting at a time of 0 and a height of 2 meters. The height of the ball follows the
drek231 [11]

Answer:

1) The height of the ball from 0 to 5  seconds are;

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2)  The correct option is;

D. -16·t² + 25·t + 1

Step-by-step explanation:

1) The equation of motion of the ball is given as follows;

H(t) = -4.9·t² + 25·t + 2

The height of the ball from 0 to 5 seconds are;

H(0) = -4.9×(0)² + 25×(0) + 2 = 2

H(1) = -4.9×(1)² + 25×(1) + 2 = 22.1

H(2) = -4.9×(2)² + 25×(2) + 2 = 32.4

H(3) = -4.9×(3)² + 25×(3) + 2 = 32.9

H(4) = -4.9×(4)² + 25×(4) + 2 = 23.6

H(5) = -4.9×(5)² + 25×(5) + 2 = 4.5

Therefore, we have;

The height of the ball are

At t = 0 second, height = 2

At t = 1 second, height h = 22.1

At t = 2 seconds, height h = 32.4

At t = 3 seconds, height h = 32.9

At t = 4 seconds, height h = 23.6

At t = 5 seconds, height h = 4.5

2) Given that the equation of the ball is that of a projectile motion, such as follows;

h = h₀ + v₀·sin(θ₀)·t - 1/2·g·t² which is equivalent to h = -1/2·g·t²+ h₀+v₀·sin(θ₀)·t

it is best represented by the quadratic equation of an upside down parabola which is option D. -16·t² + 25·t + 1

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3 years ago
Doug says that this clock shows 8:43.is he correct? Explain why or why not.
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Is this a joke because it made me laugh
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3 years ago
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Karolina [17]

Answer:

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2 years ago
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What is the minimum and maximum whole number side length for △ABC with given side lengths of 9 cm and 11 cm? Please explain why
Gwar [14]

Answer:

Step-by-step explanation:

We would assume that triangle ABC is a right angled triangle. This means that we can apply Pythagoras theorem in determining the unknown side length.

For the case of the minimum side length, we would assume that the unknown length, L is one of the shorter legs of the triangle. By applying Pythagoras theorem, it becomes

11² = 9² + L²

L² = 121 - 81 = 40

L = √40 = 6.32

For the case of the maximum side length, we would assume that the unknown length, L is one of the hypotenuse of the triangle. By applying Pythagoras theorem, it becomes

L² = 9² + 11²

L² = 81 + 121 = 202

L = √202 = 14.21

The minimum side length is 6.32 and the maximum side length is 14.21

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3 years ago
Shelby bought a 2-ounce tube of blue paint. She used 2/3 ounce to paint the water, 3/5 ounce to paint the sky, and some to paint
irina1246 [14]

Answer:

The amount of ounce used for paint is \frac{9}{15} ounce  

Step-by-step explanation:

used amount for water is \frac{2}{3} ounce.

used amount for sky is \frac{3}{5} ounce.

thus, the total amount used is \frac{2}{3} +  \frac{3}{5}

= \frac{19}{15}

Thus the total unused amount is 2 - \frac{19}{15} = \frac{11}{15} .

But the remaining amount is \frac{2}{15}.

Thus used amount for flag is \frac{11}{15} - \frac{2}{15}

= \frac{9}{15}

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