PLEASE HELP QUICK. The Hiking Club plans to go camping in a State park where the probability of rain on

What is the slope in the equation: y=4x+3

Lyrx [107]

Answer:

4

Step-by-step explanation:

The slope is 4 because the x part of the equation is the slope.

3 0

2 years ago

If ∠4 = 98°, then ∠3 = ? look at image

Ksju [112]

Step-by-step explanation:

∠4 = ∠1 & ∠2 = ∠3

360 - 2(98) = ∠2 + ∠3

∠3 = 164/2

∠3 = 82°

6 0

2 years ago

Read 2 more answers

Multiple Choice

mihalych1998 [28]

Answer:

D is correct

Step-by-step explanation:

A graph of that sort will make a perfectly mirrored "V" shape, and with no offset, the bottom point will be on the y axis. This means that the first equation will intercept the y axis at zero, the second will intercept the y-axis at -15.

"A" may seem correct also, as the second graph will intercept the x-axis at -15, but it is not complete, as it will intercept that axis at +15 as well.

7 0

2 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:

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