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1 year ago

I need to know The value of f(1/2)

Mathematics

1 answer:

Lostsunrise [7]1 year ago

7 0

$\huge\boxed{\textsf{a.)}\ 10}$

Substitute $\frac{1}{2}$ for $x$ .

$f\left(\frac{1}{2}\right)=5\cdot4^\frac{1}{2}$

Use $a^\frac{m}{n}=\sqrt[n]{a^m}$ .

$\begin{aligned}f\left(\frac{1}{2}\right)&=5\cdot\sqrt[2]{4^1}\\&=5\cdot\sqrt[2]{4}\\&=5\cdot2\\&=\boxed{10}\end{aligned}$

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The quadratic regression equation that models these data is y = –0.34x2 + 4.43x + 3.46. Using the quadratic regression equation,

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Answer:

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Step-by-step explanation:

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3 0

3 years ago

Read 2 more answers

When running a 100 meter race Bill reaches his maximum speed when he is 45 meters from the starting line and 5 seconds have elap

Mkey [24]

Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

b. 24 meters

c. 57 meters

<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

Speed Bill maintains after 5 seconds = The maximum speed

Time at which Bill is 85 meters from the starting line = 10 seconds after start

a. Required;

Bill's maximum speed in meters.

Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, t = 10 s - 5 s = 5 s

$Speed = \mathbf{\dfrac{Distance}{Time}}$

Therefore;

$Bill's \ maximum \ speed = \dfrac{40 \ m}{5 \ s} = \underline{8 \ m/s}$

b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance, s = Speed, v × Time, t

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 3 seconds = 24 meters

Therefore;

Bill's distance from the starting line 8 seconds after start is 45 meters + 25 meters = 69 meters

c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = 57 meters

Learn more about distance, speed, time, relationship here:

https://brainly.in/question/49075584

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Answer:

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