What is the solution to this equation?

Use the expression below:

skelet666 [1.2K]

Answer:

Part A: 8b -12c -16

Part B: 4(2b - 3c -4)

Part C: 36

Step-by-step explanation:

Part A:

5b + 3c - 20 + 3b - 15c + 4

= 8b -12c -16

Part B:

Factorising 8b -12c -16

= 4(2b - 3c -4)

Part C:

= 4[2(2)-3(-3)-4)]

= 4[4+9-4]

= 4[9]

= 36

7 0

3 years ago

A number n is increased by 8. If the cube root of that results equals -0.5, what is the value of n?

jek_recluse [69]

The cube root of (n increased by 8), or ∛(n+8), is -0.5, or -1/2.

∛(n+8) = -1/2. To solve this for n, cube both sides, obtaining n+8 = -1/8.

Eliminate the fraction by mult. all three terms by 8: 8n + 64 = -1

Solving for n: 8n = -65, so that n = -65/8.

8 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:

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