Who created the natural right sphilosophy

Replacing x with 0.75 in x 2 will give you an answer of 0.5625 TRUE OR FALSE.

Alina [70]

Answer

TRUE 

MARK AS BRAINLIEST 

5 0

3 years ago

I will give brainlist to the most helpful answer to my question

sdas [7]

Answer:

See below

Step-by-step explanation:

Both horses travel 0 miles in 0 minutes. We can see this on the graph where both lines start at the 0 in the bottom left corner. For the purpose of writing the equations this also shows us that the y-intercept is 0. In a slope-intercept equation, y=mx+b, that number is the b. b is zero in both equations, so we don't need to write anything for that.

For horse A, we can see on the graph that at 4 minutes, horse A has traveled 1 mile. Also, confirming this rate, at 8 minutes, it went 2 miles. This will help us find the rate. The rate will be the number we fill in for the m in the y=mx+b equation. Horse A goes 1mile every 4 minutes. That is a rate of 1/4 miles per minute. So Horse A's equation will be

y = (1/4)x You can make it more *intuitive* possibly by using m for miles and t for time instead, like this:

m = (1/4)t

Horse B is a little bit faster, and you can see this bc the line is a little bit steeper. It goes 2 miles in 5 minutes (confirm you can see it goes 4 miles in 10 minutes)

So Horse B's equation is

y = (2/5)x

or miles = (2/5)time

Mathematically, the equations are the same whether you use x,y or m,t

If Horse A runs for 12 minutes then it will run

miles = (1/4)minutes

miles = (1/4)(12)

miles = 3

If Horse B runs for 12 minutes, then it will run

miles = (2/5)minutes

miles = (2/5)12

miles = 4.8

7 0

2 years ago

Last one, please help!: :)

zalisa [80]

Answer:

I think A is right....

let see

7 0

2 years ago

Please help need answer please

DerKrebs [107]

it would be c 9/25 and 9/25

7 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$