Ask question

Ask question

All categories

3 years ago

11

A crane is lowering a concrete block from a height of 270 feet above the ground at a constant rate of 2.5 feet per second. Which

function can be used to determine h, the height , in feet above the ground of the concrete block after “s” seconds ?

1 answer:

iogann1982 [59]3 years ago

7 0

The function would be h = 270 - 2.5s, where h is the height (in feet) and s it the time (in seconds) that has passed.

The 270 represents the initial value, which is given. We are told that the initial height of the block is 270 feet.

2.5s represents the feet that it has descended. We know the block is lowered at a rate of 2.5 feet per second. Multiplying this by s, the time, will give us how much it has descended.

You might be interested in

The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

Thepotemich [5.8K]

Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

8 0

3 years ago

Is 30/12 and 40/16 equivalent?

Flura [38]

30/12 in simplest form is 5/2 (cause 6 * 5 = 30 and 6 * 2 = 12)
40/16 in simplest form is 5/2 (cause 8 * 2 = 16 and 8 * 5 =40)
so yes they're equivalent;
30/12 = 40/16
Hope that helps :D have a nice day!

5 0

3 years ago

Read 2 more answers

Point G is the centroid of the right △ABC with m∠C=90° and m∠B=30°. Find AG if CG=4 ft.

o-na [289]

Answer: $\text{Length of AG=}\frac{2\sqrt{63}}{3}$

Explanation:

Please follow the diagram in attachment.

As we know median from vertex C to hypotenuse is CM

$\therefore CM=\frac{1}{2}AB$

We are given length of CG=4

Median divide by centroid 2:1

CG:GM=2:1

Where, CG=4

$\therefore GM=2$ ft

Length of CM=4+2= 6 ft

$\therefore CM=\frac{1}{2}AB\Rightarrow AB=12$

In $\triangle ABC, \angle C=90^0$

Using trigonometry ratio identities

$AC=AB\sin 30^0\Rightarrow AC=6$ ft

$BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3}$ ft

$CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3}$ ft

In $\triangle CAN, \angle C=90^0$

Using pythagoreous theorem

$AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}$

Length of AG=2/3 AN

$\text{Length of AG=}\frac{2\sqrt{63}}{3}$ ft

5 0

3 years ago

On a coordinate plane, a line is drawn from point a to point b. point a is at (negative 8, negative 13) and point b is at (4, 11

vladimir2022 [97]

The coordinate for point P which is one-third the length of the line segment from a to b is (-4, -5)

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Point a is at (-8, -13) and point b is at (4, 11). Hence the coordinates for point p(x, y) which is one-third the length of the line segment from a to b is:

$x=\frac{1}{3}(4-(-8))+(-8)=-4\\ \\y=\frac{1}{3}(11-(-13))+(-13)=-5$

The coordinate for point P which is one-third the length of the line segment from a to b is (-4, -5)

Find out more on equation at: brainly.com/question/2972832

#SPJ4

7 0

2 years ago

Product FIFO LIFO

Julli [10]

Answer:

The null hypo would be the difference is greater than 0

so the alternate hypothesis would be less than or equal to 0

H1: µd ≤ 0

8 0

3 years ago