Find the equation of the line Use exact numbersy= x+

Mathematics

1 answer:

guapka [62]2 years ago

7 0

Answer:

I can't you didn't give the names of points on the line

Step-by-step explanation:

You might be interested in

Identify the slope of the line for the equation y = 10x + 3.

Mrrafil [7]

Answer:

D) 10

Step-by-step explanation:

The slope of the line y=10x+3 is 10 because in the slope form y=mxtb m represents the slope. In this case 10 is m, therefore the slope is 10. This mean the slope rises 10 units and runs 1 unit since slope is also known as rise/run.

8 0

3 years ago

Find the center-radius form of the equation of the circle described. Then graph the circle. Center (0,0), radius 8

Pani-rosa [81]

If the center is (0, 0), then there is no side to side or up or down motion. The center is at the origin. The standard form of a circle is (x-h)^2 + (y-k)^2 = r^2. Your h and your k are both 0's, so just fill in and square the radius:
x^2 + y^2 = 64

7 0

3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.