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2 years ago

6

4. What is the domain? Explain.

1 answer:

olasank [31]2 years ago

5 0

Answer:

The domain is (0,5) In the given pair...

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Marco starts getting ready for soccer practice at 3:55 P.M. if it takes him 15 minutes to get ready, what time is he ready to le

IceJOKER [234]

Answer:

4:10 PM

Step-by-step explanation:

15 minutes can be slit into 5 minutes and 10 minutes

There are 60 minutes in an hour!

3:55 + 0:05 = 4:00

4:00 + 0:10 = 4:10

8 0

2 years ago

Two cyclists, Alan and Brian, are racing around oval track of length 450m on the same direction simultaneously from the same poi

skelet666 [1.2K]

Answer:

Alan: 200 m/min

Brian: 150 m/min

Step-by-step explanation:

Given : Two cyclists, Alan and Brian, are racing around oval track of length 450m on the same direction simultaneously from the same point. Alan races around the track in 45 seconds before Brian does and overtakes him every 9 minutes.

To find : What are their rates, in meters per minute?

Solution :

Let n represent the number of laps that Alan completes in 9 minutes.

Then n-1 is the number of laps Brian completes.

45 seconds = 3/4 minutes.

The difference in their lap times in minutes per lap is

$\frac{9}{(n -1)}-\frac{9}{n} =\frac{3}{4}$

Solving the equation we get,

$\frac{9n-9n+9}{n(n -1)} =\frac{3}{4}$

$9\times 4=3\times n(n -1)$

$36=3\times n(n -1)$

$n^2-n-12=0$

$n^2-4n+3n-12=0$

$n(n-4)+3(n-4)=0$

$(n-4)(n+3)=0$

$n=4,-3$

Neglecting n=-3

So, n=4

Then Alan's speed in m/min is

$S=\frac{D}{T}$

$S_a=\frac{4\times 450}{9}$

$S_a=200 m/min$

Brian completes 3 laps in that 9-minute time, so his rate is

$S_b=\frac{3\times 450}{9}$

$S_b=150 m/min$

Therefore, Alan: 200 m/min

Brian: 150 m/min

4 0

3 years ago

What is the equation of the line that passes through the point (-2, 7) and has a slope of zero

lorasvet [3.4K]

Answer:

y = 7 is the equation of the line that passes through the point ( -2, 7 ) and has a slope of zero.

Step-by-step explanation:

Given:

Let,

A ≡ ( x1 , y1 ) ≡ ( -2, 7 )

Slope = m = 0

To Find :

Equation of Line:

Solution:

Formula for , equation of a line passing through a point ( x1 , y1 ) and having a slope m is given by

$(y - y_{1})=m(x-x_{1})$

Now substituting the values of x1 = -2 and y1 = 7 and slope m = 0 we get,

$y-7=0\times(x--2) \\y-7=0\times (x+2)\\y-7=0\\\therefore y=7$

Which is the required equation of a line passing through the point ( -2, 7 ) and slope zero

6 0

3 years ago

Can anyone please help me with this!!! :(

UkoKoshka [18]

Answer:

ano ba gagawin

parang Ang Dali pero ano gagawin dyan

3 0

2 years ago

Read 2 more answers

Find the point (,) on the curve =8 that is closest to the point (3,0). [To do this, first find the distance function between (,)

ELEN [110]

Question:

Find the point (,) on the curve $y = \sqrt x$ that is closest to the point (3,0).

[To do this, first find the distance function between (,) and (3,0) and minimize it.]

Answer:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

Step-by-step explanation:

$y = \sqrt x$ can be represented as: $(x,y)$

Substitute $\sqrt x$ for $y$

$(x,y) = (x,\sqrt x)$

So, next:

Calculate the distance between $(x,\sqrt x)$ and $(3,0)$

Distance is calculated as:

$d = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

So:

$d = \sqrt{(x-3)^2 + (\sqrt x - 0)^2}$

$d = \sqrt{(x-3)^2 + (\sqrt x)^2}$

Evaluate all exponents

$d = \sqrt{x^2 - 6x +9 + x}$

Rewrite as:

$d = \sqrt{x^2 + x- 6x +9 }$

$d = \sqrt{x^2 - 5x +9 }$

Differentiate using chain rule:

Let

$u = x^2 - 5x +9$

$\frac{du}{dx} = 2x - 5$

So:

$d = \sqrt u$

$d = u^\frac{1}{2}$

$\frac{dd}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

Chain Rule:

$d' = \frac{du}{dx} * \frac{dd}{du}$

$d' = (2x-5) * \frac{1}{2}u^{-\frac{1}{2}}$

$d' = (2x - 5) * \frac{1}{2u^{\frac{1}{2}}}$

$d' = \frac{2x - 5}{2\sqrt u}$

Substitute: $u = x^2 - 5x +9$

$d' = \frac{2x - 5}{2\sqrt{x^2 - 5x + 9}}$

Next, is to minimize (by equating d' to 0)

$\frac{2x - 5}{2\sqrt{x^2 - 5x + 9}} = 0$

Cross Multiply

$2x - 5 = 0$

Solve for x

$2x =5$

$x = \frac{5}{2}$

Substitute $x = \frac{5}{2}$ in $y = \sqrt x$

$y = \sqrt{\frac{5}{2}}$

Split

$y = \frac{\sqrt 5}{\sqrt 2}$

Rationalize

$y = \frac{\sqrt 5}{\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$y = \frac{\sqrt {10}}{\sqrt 4}$

$y = \frac{\sqrt {10}}{2}$

Hence:

$(x,y) = (\frac{5}{2},\frac{\sqrt{10}}{2}})$

3 0

3 years ago