PLEASE HELP ME WITH THIS QUESTION.

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

5 0

The answer is B

-b/2(a) = x
That is the time when the ball hits the ground

Gwar [14]3 years ago

5 0

Answer:

The ball hits the ground after 7.6 sec.

Step-by-step explanation:

Realize that h = 0 when the rocket hits the ground. Thus, we set h(t) = y = to 0 and solve for time (t):

y = 0 = h(t) = -16t^2 + 113t + 65.

Application of the quadratic formula is the easiest approach here. Note that a = -16, b = 113 and c = 65.

The discriminant is b^2-4ac, or, in this case, 113^2 - 4(-16)(65) = 16929.

Because the discriminant is positive, we confirm that this equation has two real, unequal roots.

The time values are as follows:

-113 ± √16929

t = --------------------- = -17.11/ (-32) sec, which we must reject

-32

because time in

this situation may not be (-).

The other root is:

-113 ± √16929

t = --------------------- = 7.6 sec

-32

The ball hits the ground after 7.6 sec.

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A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

Read 2 more answers

I will gove ppints to anybody who answers this because you will deserve it thank you

kkurt [141]

$\left(4^4\right)^2=4^8=65536\\
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