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3 years ago

15

A water balloon is 5 feet above the ground when Sally launches it into the air. Use the quadratic equation 0 = -t^2 + 4t + 5 to

find how much time, t, it takes for the water balloon to reach the ground.

2 answers:

Alla [95]3 years ago

8 0

Answer:

<u><em>5 seconds </em></u>

Step-by-step explanation:

ttm/imagine math

rodikova [14]3 years ago

7 0

Answer:

$t=5\ sec$

Step-by-step explanation:

we have

$-t^{2} +4t+5=0$

Solve the quadratic equation to find the zero's or x-intercepts

Remember that

The x-intercepts are the values of x when the value of y is equal to zero ( water balloon reach the ground)

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-t^{2} +4t+5=0$

so

$a=-1\\b=4\\c=5$

substitute in the formula

$t=\frac{-4(+/-)\sqrt{4^{2}-4(-1)(5)}} {2(-1)}$

$t=\frac{-4(+/-)\sqrt{36}} {-2}$

$t=\frac{-4(+/-)6} {-2}$

$t=\frac{-4(+)6} {-2}=-1$

$t=\frac{-4(-)6} {-2}=5$

therefore

the solution is $t=5\ sec$

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Step-by-step explanation:

For this case we can use the notable angls given on the picture attached.

Part a

For this case we can use the fact that $sin (\pi/3) = \frac{\sqrt{3}}{2}$

And for this case since we ar einterested on $-\frac{\pi}{3}$ and we know that the if we are below the y axis the sine would be negative then:

$sin (-\pi/3) = -\frac{\sqrt{3}}{2}$

Part b

From definition we can use the fact that $tan x= \frac{sin x}{cos x}$ and we got this:

$tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}$

We can use the notabl angle $\pi/4$ and we know that :

$sin (\pi/4) = cos(\pi/4) = \frac{\sqrt{2}}{2}$

Then we know that $5\pi/4$ correspond to 225 degrees and that correspond to the III quadrant, and we know that the sine and cosine are negative on this quadrant. So then we have this:

$tan (5\pi/4) = \frac{sin(5\pi/4)}{cos(5\pi/4)}= \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$

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