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Dovator [93]
3 years ago
8

Graph the function f(x) = 1/2(2)^x

Mathematics
1 answer:
xxMikexx [17]3 years ago
3 0

Answer; Down

Step-by-step explanation:The graph of the squaring function has the shape of a parabola that opens up. Its vertex is at (0,0). The graph of F(x) = (1/2)x^2 has the same shape, but is compressed vertically by a factor of (1/2).

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Kristina and Darrin are running around a circular track with the same angular speed. Darrin is running on the inside lane which
Nataliya [291]

Answer:

10 ft/s

Step-by-step explanation:

If Kristina is running at a rate of 11 ft/s and on a circular track with radius of 44 ft, then her angular speed is:

\omega_K = \frac{v_K}{r_K} = \frac{11}{44} = 0.25 rad/s

And so is Darrin's angular speed. That means his rate is

v_D = \omega_D * r_D = 0.25 * 40 = 10 ft/s

7 0
3 years ago
Mr. Richard owns an orchard that has a rectagular fence . The orchard is 36 yards long and 18 yards wide . if he walks across th
Sergio [31]

Answer:

40.2 Yards

Step-by-step explanation:

To find the diagonal length of the orchard, we need to use the Pythagorean Theorem. The formula of the Pythagorean Theorem is:

c²= a² + b²

a = 36 Yards

b = 18 Yards

Now let's plug in our values to the formula.

c² = a² + b²

c² = 36² + 18²

c² = 1296 + 324

c² = 1620

Now to find the value of c we need to get the square root of both sides.

\sqrt{c^{2}}=\sqrt{1620}

c = 40.2

So the diagonal length of Mr. Richard's orchard is 40.2 Yards.

7 0
4 years ago
10 plus what equals negative 60
Fiesta28 [93]

10 + -70

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Can someone help me solve this question
Norma-Jean [14]

Answer:

Step-by-step explanation:

Choosing a white pen from box a

8 0
3 years ago
A ball is thrown into the air with an upward velocity of 38 ft/s. Its height h(t) in feet after t seconds is given by the functi
Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

In a quadratic modeling free fall, h(t) = -16t² + v₀t + h, h(t) is the height of the object AFTER the fall while h is the initial height of the object.  To answer a:

a. The initial height of the object is 7 feet

To find the height, h(t), after 1.5 seconds for b., evaluate the quadratic at h(1.5):

h(1.5) = -16(1.5)² + 38(1.5) + 7 so

b. h(1.5) = 29.5 feet

In order to determine the max height of the object, we need to put this quadratic into vertex form.  This is an upside down (negative) parabola, so the vertex represents a max height.  It is at the vertex values of (h, k) that we will find the answers to both c and d.  To put this into vertex form, we need to complete the square on the quadratic.  Do this by first setting the quadratic equal to 0 then moving over the constant to get:

-16t² + 38t = -7

The first rule for completing the square is that leading coefficient has to be a positive 1.  Ours is a -16, so we have to factor it out:

-16(t² - 2.375t) = -7

The next rule for completing the square is to take half the linear term, square it, and then add that to both sides.  Our linear term is 2.375.  Half of that is 1.1875, and 1.1875 squared is 1.41015625.  BUT we cannot forget about that -16 sitting out front of those parenthesis.  It is a multiplier.  That means that we did not just add in 1.41015625, we added in -16(1.41015625):

-16(t² - 2.375t + 1.41015625) = -7 - 22.5625

The reason we do this is to create a perfect square binomial on the left.  Writing the left side in terms of our binomial and at the same time simplifying the right:

-16(t - 1.1875)² = -29.5625

We finish it off by adding 29.5625 to both sides and setting the vertex form of the quadratic back equal to y:

y = -16(t - 1.1875)² + 29.5625

From this we can ascertain that the vertex is located at (1.1875, 29.5625)

The answer to c is found in the h coordinate of the vertex which 1.1875.  

c. 1.1875 seconds to reach its max height

The answer to d is found in the k coordinate of the vertex which is 29.5625

d. The max height is 29.5625 feet

For e, we will look back to the original function.  We determined that h(t) is the height of the object AFTER it falls.  If the object falls to the ground, it goes to follow that the height of the object while it is on the ground is 0.  We sub in 0 for h(t) and factor to find the times at which the object hits the ground.

Plugging our values into the quadratic formula gives us the times of -.017 seconds and 2.5 seconds.  Since we know that time can't EVER be negative, the object hits the ground 2.5 seconds after it was dropped.

Good luck with your quadratics!

6 0
3 years ago
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