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3 years ago

The equation of a line perpendicular to y = 3 and passes through the point (24, -56).

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

3 0

Answer:

x = 24

Step-by-step explanation:

The given line y = 3 is a horizontal line with slope zero (0).

We wish to find the equation of a line that is perpendicular to y = 3. Such a line would be a vertical one. Vertical lines do not have slopes defined (due to division by zero).

Thus the general form of the equation of this new line is x = c.

This new line passes through (24, -56). Thus, x must be 24: x = 24

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A rectangle as the width of 9 units and length of 40 units. What is the length of the diagonal?

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Answer:

the Diagonal is 41 units

Step-by-step explanation:

Since the sides of a rectangle, to be a rectangle, meet at 90° we use Pythagorean Theorem a² + b² = c²

Width and Length are the legs and the diagonal is the hypotenuse

9² + 40² = c²

81 + 1600 = c²

1681 = c²

√1681 = √c²

41 = c

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3 years ago

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3 years ago

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What is the value of f(-4) in the piecewise function f(x)= -x-5 for -5<= x <= -2; -x^2+1 for -2<=x<=2; (x-3)^2+2 for

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We first need to pick the correct piece to use. In this case, we will use f(x) = -x-5 because -5<=-4.

Then, we use direct substitution:
f(-4) = -(-4) - 5
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Which pair of angles are an example of alternate exterior angles?

9966 [12]

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c. is the answer becuase since it is exterior they are talking about the outside of the parellel lines.

Step-by-step explanation:

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A soccer ball is kicked from 3 feet above the ground. the height,h, in feet of the ball after t seconds is given by the function

dexar [7]

Here's our equation.

$h=-16t+64t+3$

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

$0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}$

$= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}$

So the ball will return to the ground at the positive value of $\boxed{\frac{8+\sqrt{67}}{4}}$ seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

$\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2$

$h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}$

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3 years ago