Which of the following numbers are greater than or equal to 4/7

After 2 week he has used 8 1/2 cups of dog food after 5 week he has used 21 1/4 cups construct a function in the form y =mx+ to

gladu [14]

Answer:

y = 4.25x

Step-by-step explanation:

Given that:

After 2 week he has used 8 1/2 cups of dog food after 5 week he has used 21 1/4 cups, we can write this in coordinate form (x, y) where;

y is the amount of dog food

x is the time in weeks

The coordinates are (2, 8 1/2) and (5, 21 1/4)

get the slope m:

m = y2-y1/x2-x1

m = (21 1/4 - 8 1/2)/5-2

m = (85/4-17/2)/3

m = (85-34/4)/3

m = 51/12

m = 4.25

Get the intercept c:

Substitute any of the points say (2, 17/2) and m = 51/12 into the equation

y = mx+c

17/2 = 51/12(2) + c

17/2 = 51/6 + c

c = 17/2 - 51/6

c = 51-51/6

c = 0

Get the required equation:

substitute m - 4.25 and c = 0 into y = mx+c

y = 4.25x + 0

Hence the required equation is y = 4.25x

8 0

2 years ago

If aerts305 is 16 and his friend is 15 who’s older

Angelina_Jolie [31]

Answer:

aerts is older by 1 year

Step-by-step explanation:

5 0

2 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.

7 0

3 years ago

What is the value of 5.4?

Galina-37 [17]

The value is .54 hope this helped

5 0

2 years ago

Sort the cities from closest to farthest from sea level. Please help me

dimaraw [331]

I think
J,BB,S

(I shortened name to letter)

5 0

2 years ago