I need Answer Immediately!!!!!!

Solve for m in D m/v, if D= 5.1 and v = 0.3 the / is for a fraction.

laila [671]

5.1 = m/0.3
1.53 = m
Hope this helps!

8 0

3 years ago

Solve the linear equation - 7(x + 2) = -3(x - 5) for x

melisa1 [442]

x= −29/4

Answer:

Step-by-step explanation:

−7(x+2)=−3(x−5)

Step 1: Simplify both sides of the equation.

−7(x+2)=−3(x−5)

(−7)(x)+(−7)(2)=(−3)(x)+(−3)(−5)(Distribute)

−7x+−14=−3x+15

−7x−14=−3x+15

Step 2: Add 3x to both sides.

−7x−14+3x=−3x+15+3x

−4x−14=15

Step 3: Add 14 to both sides.

−4x−14+14=15+14

−4x=29

Step 4: Divide both sides by -4.

−4x −4

x=29/−4

x= −29/4

3 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:

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