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JulijaS [17]
3 years ago
13

Someone please explain (:

Mathematics
2 answers:
GaryK [48]3 years ago
7 0

Answer:

a.) 1.38 seconds

b.) 17.59ft

Step-by-step explanation:

h(t) = -16t^2 + 22.08t + 6

if we were to graph this, the vertex of the function would be the point, which if we substituted into the function would give us the maximum height.

to find the vertex, since we are dealing with something called "quadratic form" ax^2+bx+c, we can use a formula to find the vertex

-b/2a

b=22.08

a=-16

-22.08/-16, we get 1.38 when the minuses cancel out. since our x is time, it will be 1.38 seconds

now since the vertex is 1.38, we can substitute 1.38 into the function to find the maximum height.

h(1.38)= -16(1.38)^2 + 22.08t + 6 -----> is maximum height.

approximately = 17.59ft -------> calculator used, and rounded to 2 significant figures.

for c the time can be equal to (69+sqrt(8511))/100, as the negative version would be incompatible since we are talking about time. or if you wanted a rounded decimal, approx 1.62 seconds.

vichka [17]3 years ago
3 0

Answer:

a) h(t)=-16(t-0.69)^2)+13.62

b)The maximum height is 13.62 feet.

c) 1.84 seconds

Step-by-step explanation:

The function that models the height in feet of the hammer above the ground is

h(t)=-16t^2+22.08t+6

Factor  -16 as shown;

h(t)=-16(t^2-1.38t)+6

Add and subtract  the square of half the coefficient of t.

h(t)=-16(t^2-1.38t+(-0.69)^2)--16(-0.69)^2+6

h(t)=-16(t^2-1.38t+(-0.69)^2)--16(-0.69)^2+6

Rewrite the first three terms as a perfect square trinomial;

h(t)=-16(t-0.69)^2)+13.62

This function is now in the form;

h(t)=a(t-h)^2)+k

where (h,k)=(0.69,13.62)

This implies that; the hammer reached its maximum height at t=0.69 seconds.

b) The maximum height is the y-value of the vertex.

The maximum height is 13.62 feet.

c) To answer this question we need to find the difference between the  x-intercepts.

h(t)=-16t^2+22.08t+6

a=-16,b=22.08,c=6

t_2-t_1=\sqrt{(t_1+t_2)^2-4t_1t_2}

t_2-t_1=\sqrt{(\frac{-b}{a})^2-4(\frac{c}{a})}

t_2-t_1=\sqrt{(\frac{-22}{-16})^2-4(\frac{-6}{-16})  }=1.84

Hence the hammer was in the air for 1.84 seconds

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Here is the full question:

Triangle ABC is a right-triangle and sin(53o) = StartFraction 4 Over x EndFraction. Solve for x & round to the nearest whole number.

Triangle A B C is shown. Angle A C B is a right angle and angle B A C is 53 degrees. The length of B C is 4 centimeters, the length of A C is y, and the length of hypotenuse A B is x.

Which equation correctly use the value of x to represents the cosine of angle A?

cos(53o) = StartFraction 4 Over x EndFraction cos(53o) = StartFraction y Over 5 EndFraction cos(53o) = StartFraction x Over 4 EndFraction cos(53o) = StartFraction 5 Over y EndFraction

Answer:

cos(53°) = StartFraction y Over 5 EndFraction

Step-by-step explanation:

From the information above:

The sketch of the triangle is drawn and attached in the diagram below.

To find x using the sine rule,

We know that:

Sin \theta = \dfrac{opposite}{hypothenuse}

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Making x the subject of the formula:

x = \dfrac{4}{Sin \ 53^0}

x = \dfrac{4}{0.7986}

x \simeq 5

Now, the equation that correctly uses the value of x can be determined by finding the cosine of ∠ A.

Recall that:

Cos \theta = \dfrac{adjacent}{hypothenuse}

Cos \ 53^0 = \dfrac{y}{x}

where; x = 5

Cos \ 53^0 = \dfrac{y}{5}

Thus, the equation that correctly uses the value of x to represent the cosine of angle A is:

cos(53°) = StartFraction y Over 5 EndFraction

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