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4 years ago

10

If 4/5 of the population of a certain town is considered to be middle class and the population of the town is 2,000,how many peo

ple are considered middle class

1 answer:

kicyunya [14]4 years ago

7 0

1,600 people are considered middle class.

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Simplify this expression leaving your answer in index form

kenny6666 [7]

Answer:

solution is 9a^16

Step-by-step explanation:

(6)^2= 36

(a^5)^2= a^25

36a^25/(9a^9)

(36/9)= 4

(a^25)/(a^9)= a^(25-9)= a^16

3 0

3 years ago

A car is travelling at exactly 6 miles per hour and accelerates at a constant rate to exactly 65 miles per hour.

vichka [17]

Answer:

$14878.04878miles/hours^2$

Step-by-step explanation:

Let's find a solution by understanding the following:

The acceleration rate is defined as the change of velocity within a time interval, which can be written as:

$A=(Vf-Vi)/T$ where:

A=acceleration rate

Vf=final velocity

Vi=initial velocity

T=time required for passing from Vi to Vf.

Using the problem's data we have:

Vf=65miles/hour

Vi=6miles/hour

T=14.8seconds

Using the acceleration rate equation we have:

$A=(65miles/hour - 6miles/hour)/14.8seconds$ , but look that velocities use 'hours' unit while 'T' uses 'seconds'.

So we need to transform 14.8seconds into Xhours, as follows:

$X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)$

$X=0.0041hours$

Using X=0.0041hours in the previous equation instead of 14.8seconds we have:

$A=(65miles/hour - 6miles/hour)/0.0041hours$

$A=(61miles/hour)/0.0041hours$

$A=(61miles)/(hour*0.0041hours)$

$A=61miles/0.0041hours^2$

$A=14878.04878miles/hours^2$

In conclusion, the acceleration rate is $14878.04878miles/hours^2$

7 0

3 years ago

Solve l 3x-7 l = 2

Lelechka [254]

Answer:

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

$\implies x=\frac{7\pm 2}{3}$

$x=\frac{7+2}{3}\text{ or }x=\frac{7-2}{3}$

$\implies x = 3\text{ or }x=\frac{5}{3}$

2) l 2x-5 l ≤ 8

⇒ 2x-5 ≤ ±8

⇒ 2x ≤ 5 ± 8

$\implies x\leq \frac{5\pm 8}{2}$

$x\leq \frac{5+8}{2}\text{ or }x\leq \frac{5-8}{2}$

$\implies x \leq \frac{13}{2}\text{ or }x\leq-\frac{3}{2}$

8 0

3 years ago

Read 2 more answers

If a phone card is used to make a long distance phone call, you are charged $0.50 per call plus an additional $0.31 per minute.

Nikitich [7]

Part A: c for cost. c=0.31m+0.5

0.31m is the cost per minute. 0.5 is cost per call.

Part B: 0.31m+0.5=5.15 to solve we must rearrange.

subtract 0.5 from each side giving us 0.31m=4.85

divide by 0.31 giving us m=15.65

7 0

3 years ago

One tank is filling at a rate of 5/8 gallon per 7/10 hour. A second tank is filling at a rate of 5/9 gallon per 2/3 hour. Which

anzhelika [568]

First recognize that the unit rate we're finding is gallons per hour, or $\frac{g}{h}$ .

rate for Tank 1 = x gallons per hour = <em>a</em> gallons / <em>b</em> hours

$T_{1}=\frac{\frac{5}{8}}{\frac{7}{10}}\\\\T_{1}=\frac{5}{8}*\frac{10}{7}\\\\T_{1}=\frac{50}{56}\\\\\\Fill\ rate\ of\ Tank\ 1=\frac{25}{28}$

rate for Tank 2 = y gallons per hour = <em>c</em> gallons / <em>d</em> hours

$T_{2}=\frac{\frac{5}{9}}{\frac{2}{3}}\\\\T_{2}=\frac{5}{9}*\frac{3}{2}\\\\T_{2}=\frac{15}{18}\\\\T_{2}=\frac{15}{18}\\\\\\Fill\ rate\ of\ Tank\ 2=\frac{5}{6}$

3 0

3 years ago