8-2b=-2/3(12b+6) solve

2 answers:

7 0

We have to mutiply factor -2/3 times every agument from the parenthesis.
$8-2b=- \frac{2*12b}{3}- \frac{2*6}{3} \\ 8-2b= -8b-4 \\$ /+8b
8+6b=-4 /-8
6b=-12 /:6
b=-2 - its the answer

pishuonlain [190]2 years ago

6 0

$8-2b=-\frac { 2 }{ 3 } \left( 12b+6 \right) \\ \\ 8-2b=-\frac { 2 }{ 3 } \cdot 12b-\frac { 2 }{ 3 } \cdot 6\\ \\ 8-2b=-\frac { 24 }{ 3 } b-\frac { 12 }{ 3 } \\ \\ 8-2b=-8b-4\\ \\ -2b+8b=-4-8\\ \\ 6b=-12\\ \\ b=\frac { \left( -12 \right) }{ 6 } \\ \\ \therefore \quad b=-2$

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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:

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