What is 9/10 of 7 inches

A basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modele

aev [14]

THIS IS THE COMPLETE QUESTION

basketball player is shooting a basketball toward the net. The height, in feet, of the ball t seconds after the shot is modeled by the equation h = 6 + 30t – 16t2. Two-tenths of a second after the shot is launched, an opposing player leaps up to block the shot. The height of the shot blocker’s outstretched hands t seconds after he leaps is modeled by the equation h = 9 + 25t – 16t2. If the ball reaches the net 1.7 seconds after the shooter launches it, does the leaping player block the shot?

Yes, exactly 0.6 seconds after the shot is launched.

Yes, between 0.64 seconds and 0.65 seconds after the shot is launched

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

No, the shot is not blocked.

Answer:

Yes, between 0.84 seconds and 0.85 seconds after the shot is launched.

Given that the

equation for shot block height h = 9 + 25t – 16t2.

equation for ball height h = 6 + 30t – 16t2.

From the question ,it can be deducted that the shot is made before two tenths of a second or 0.2seconds

CHECK THE ATTACHMENT TO COMPLETE THE DETAILED SOLUTION

5 0

3 years ago

Read 2 more answers

If the measure of one angle in a pair of is n, the<br> The measure of the other angle is 180 – n.

olchik [2.2K]

This kinda doesn’t make sense but maybe 80, 80, 20

8 0

3 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

7 0

3 years ago

Circle C with center at (−4, 6) and radius 2 is similar to circle D with center at (6, −2) and radius 4. Below is an incorrect i

BlackZzzverrR [31]

We are told that circle C has center (-4, 6) and a radius of 2.

We are told that circle D has center (6, -2) and a radius of 4.

If we move circle C's center ten units to the right and eight units down, the new center would be at (-4 + 10), (6 - 8) = (6, -2). So step 1 in the informal proof checks out - the centers are the same (which is the definition of concentric) and the shifts are right.

Let's look at our circles. Circle C has a radius of 2 and is inside circle D, whose radius is 4. Between Circle C and Circle D, the radii have a 1:2 ratio, as seen below:

$\frac{1}{2} = \frac{radius--circle C}{radius--circle D}$

If we dilate circle C by a factor of 2, it means we are expanding it and doubling it. Our circle has that 1:2 ratio, and doubling both sides gives us 2:4. The second step checks out.

Translated objects (or those that you shift) can be congruent, and dilated objects are used with similarity (where you stretch and squeeze). The third step checks out.

Thus, the argument is correct and the last choice is best.

3 0

4 years ago

What is the volume of the solid?

Evgen [1.6K]

9514 1404 393

Answer:

(9√3 -3π/2) ft^3 ≈ 10.88 ft^3

Step-by-step explanation:

The area of the hexagon is given by the formula ...

A = (3/2)√3·s^2 . . . . for side length s

The area of the hexagonal face of this solid is ...

A = (3/2)√3·(2 ft)^2 = 6√3 ft^2

__

The area of the circular hole in the hexagonal face is ...

A = πr^2

The radius is half the diameter, so is r = (2 ft)/2 = 1 ft.

A = π(1 ft)^2 = π ft^2

Then the area of the "solid" part of the face of the figure is ...

A = (6√3 -π) ft^2

__

The volume is ...

V = Bh . . . . . where B is the area of the base of the prism, and h is its height

V = ((6√3 -π) ft^2)(3/2 ft) = (9√3 -3π/2) ft^3 ≈ 10.88 ft^3

8 0

3 years ago