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3 years ago

PLEASE HELP DUE TODAY WILL MARK BRAINLIEST

Mathematics

1 answer:

ki77a [65]3 years ago

8 0

Answer: y= -2x - 500

Step-by-step explanation:

Parallel lines have the same slopes but different y-intercepts. This means that the line parallel to y=-2x + 3 will have the same slope which is -2.

We will input the x and y coordinates for the given point into the formula and solve for the y-intercept to write the equation.

-100 = -2(-100) + b where b is the y-intercept

-100 = 400 + b

-400 -400

b = -500

The y intercept is -500 and the slope is -2 therefore, the equation will be

y= -2x - 500

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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:

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

Since time can only be positive we reject the $t_{2}$ solution and we keep that Ben's book took $t=4.3276$ seconds to reach the ground.

Similarly solving for Josh we obtain

$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

Whose textbook reaches the ground first and by how many seconds?

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

3 0

3 years ago

PLEASE HELP ASAP I DON'T UNDERSTAND

Ira Lisetskai [31]

The slope of a linear equation will tell you the price of each ticket. The y-intercept tells you the convenience fee.
We can find the slope by choosing two points and finding the slope.
m = (y2 - y1)/(x2 - x1)
m = (887.40 - 231.15)/(20 - 5)
m = 656.25/15
m = 43.75
Each ticket will cost $43.75

4 0

2 years ago

What is the area of this figure? Enter your answer in the box. mm²

Rainbow [258]

Answer:

$84mm^{2}$