Is 843 divisible by 10? yes no Submit

How does -5 + 5 equal 0? Explain

Katyanochek1 [597]

Answer:

Because your taking away 5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I need to know how to solve the equation 2/3x+1= -1/3x-1

butalik [34]

So x should be equal to -2 when you equal them to each other.

3 0

3 years ago

The diagnosis of a square are a.perpendicular

makkiz [27]

Answer

The correct answer is B

Step-by-step explanation:

It is B because squares don't intersect so it cant be a

They are always congruent and equal so not c or d it has to be B

6 0

3 years ago

What is the margin of error for a yes/no survey of 200 people with a proportion of 65% who answered \"yes\"?. A. ±3.37%. B. ±5.0

azamat

$\sqrt{(p*q)/n}$

where p is the number of positive outcomes (yes) and q is the number of negative outcomes (no).

$\sqrt{(0.65*(1-0.65))/200}$ = 0.0337 or 3.37%

answer is A.

6 0

4 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:

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