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Ronch [10]
1 year ago
15

A function, f, has an x-intercept at (2,0) and a y-intercept at (0.10)

Mathematics
2 answers:
aalyn [17]1 year ago
5 0

A function, f, with an x-intercept at (2,0) and a y-intercept at (0, 10) has an equation y = -5x + 10

<h3>How to find the equation of a line?</h3>

The equation of a line in slope intercept form is expressed as y =x +b

where

  • m is the slope
  • b is the y-intercept

Given a function, f, has an x-intercept at (2,0) and a y-intercept at (0, 10), the slope is expressed as:

Slope = 10/-2
Slope = -5

Substitute m = -5 and b = 10 into the formula to have y = -5x + 10

Learn more on equation of a line here: brainly.com/question/27680685

#SPJ1

valentinak56 [21]1 year ago
4 0

The equation of the function is f(x) = -5x + 10

<h3>How to determine the function equation?</h3>

The points are given as:

x-intercept at (2,0) and a y-intercept at (0,10)

A linear function is represented as:

y = mx + b

Where b is the y-intercept.

The y-intercept at (0,10) means that:

b = 10

So, we have:

y = mx + 10

The x-intercept at (2,0) means that:

x = 2, when y = 0

Substitute these values in y = mx + 10

0 = 2m + 10

Subtract 10 from both sides

2m = -10

Divide both sides by 2

m = -5

Substitute m = -5 in y = mx + 10

y = -5x + 10

Hence, the equation of the function is y = -5x + 10

Read more about linear functions at:

brainly.com/question/4025726

#SPJ1

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A spherical balloon currently has a radius of 19cm. If the radius is still growing at a rate of 5cm or minute, at what rate is a
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22670.8 cm³/min

Step-by-step explanation:

Given:

Radius of the balloon at a certain time (r) = 19 cm

Rate of growth of radius is, \frac{dr}{dt}=5\ cm/min

The rate at which the air is pumped in the balloon can be calculated by finding the rate of increase in the volume of the balloon.

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Volume of balloon is given as:

V=\frac{4}{3}\pi r^3

Now, rate of increase of volume is obtained by differentiating both sides of the equation with respect to time 't'.

Differentiating both sides with respect to time 't', we get:

\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)\\\\\frac{dV}{dt}=\frac{4\pi}{3}(3r^2)(\frac{dr}{dt})\\\\\frac{dV}{dt}=4\pi r^2(\frac{dr}{dt})

Now, plug in 19 cm for 'r', 5 cm per minute for \frac{dr}{dt} and solve for \frac{dV}{dt}. This gives,

\frac{dV}{dt}=4\pi (19 cm)^2(5\ cm/min)\\\\\frac{dV}{dt}=4\times 3.14\times 361\times 5\ cm^3/min\\\\\frac{dV}{dt}=22670.8\ cm^3/min

Therefore, the rate at which the air is being pumped into the balloon is 22670.8 cm³/min.

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Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off a
zubka84 [21]

Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a =  (v - u)/t =  (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

The distance moved by the police car in 10 s is gotten from

s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.

s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

s = 122.9 m

The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h

The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' =  17.88t' m where t' = time taken for police to catch up with speeder.

Since both distances are the same,

S' = S

17.88t' = 122.9 + 24.58t

Also, the time  taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

17.88(10 + t) = 122.9 + 24.58t

178.8 + 17.88t = 122.9 + 24.58t

17.88t - 24.58t = 122.9 - 178.8

-6.7t = -55.9

t = -55.9/-6.7

t = 8.34 s

So, t' = 10 + t

t' = 10 + 8.34

t' = 18.34 s

So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

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2 years ago
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