All categories

4 years ago

4 (b - 3 ) = 6 steps in order from first to last

Mathematics

2 answers:

Amiraneli [1.4K]4 years ago

8 0

Answer:

B= 9/2 Hope this helped <3

Step-by-step explanation:

Let's solve your equation step-by-step.

4(b−3)=6

Step 1: Simplify both sides of the equation.

4(b−3)=6

(4)(b)+(4)(−3)=6(Distribute)

4b+−12=6

4b−12=6

Step 2: Add 12 to both sides.

4b−12+12=6+12

4b=18

Step 3: Divide both sides by 4

Lorico [155]4 years ago

3 0

Answer:

b=4.5

Step-by-step explanation:

distribute

4b-12=6

add 12

4b=18

divide 4

b=4.5

You might be interested in

What is 8+2x+7x equal

Veseljchak [2.6K]

I’m pretty sure it’s 9x+8

5 0

4 years ago

Read 2 more answers

Armando got a job selling bags of peanuts at the local art festival. He is paid a fixed amount of $320 per

skelet666 [1.2K]

0.67n + 320

=(0.67 * 430) + 320

= 288.1 + 320

= $608.1

0.67n + 320 = 728.70

0.67n = 728.70 - 320

0.67n = 408.7

n = 408.7 / 0.67

n = 610 Bags of peanuts

6 0

3 years ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.