Write the equation of the line perpendicular to y = -2x -3 that goes through the point (2, -3)

Frank started out in his car travelling 45 mph. When Frank was 1/3 miles away, Daniel started out from the same point at 50 mph

Anna007 [38]

Time taken by Daniel to catch up with Frank is 4 minutes.

Step-by-step explanation:

The question is on relative speed; speed of a moving object with respect to another.

The two cars are travelling in the same direction at different speeds thus;

Relative speed = speed of Daniel - speed of Frank = 50-45= 5 mph

The time which the two meet = distance/ relative speed

Distance = 1/3 miles, which equals 0.33 miles

Time taken by Daniel to catch up with Frank is = 0.33/5 =0.066 hours = 4 minutes.

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Relative speed: brainly.com/question/10681729

Keywords: car, travelling, miles,started at the same point,catch up with

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6 0

3 years ago

Read 2 more answers

Assume lim f(x)-6 and lim g(x)-9. Compute the following limit and state the limit laws used to justify the computation.

emmasim [6.3K]

Answer:

4

Step-by-step explanation:

<h3>some relevant limit laws</h3>

lim C = C where c is a constant.

lim( f(x) + g(x)) =lim f(x) + lim g(x)

lim( f(x)g(x)) =lim f(x) * lim g(x)

lim( cg(x)) =clim g(x)

lim( f(x)/g(x)) =lim f(x) / lim g(x) if lim g(x) is not equal to zero.

lim( f(x))^2 = (lim f(x) )^2

lim square root( f(x)) = square root(lim f(x) )

$\lim_{n \to 3} g(x) = 9\\\\\lim_{n \to 3} f(x) = 6\\\\ \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10} \\\\ = \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10}\\\\= \sqrt[3]{lim_{n \to 3}f(x) \times lim_{n \to 3}g(x) + 10}\\\\= \sqrt[3]{6 \times 9 + 10}\\\\= \sqrt[3]{64}$

= 4

4 0

3 years ago

A clothing store donated a percent of every sale to hospitals. The total sales were $6,500 so the store

ivanzaharov [21]

The answer is 4%

1) work out 1%
6500/100 = 65

2)work out how many times 65 goes into 260
260/65= 4

8 0

3 years ago

Helppppp me please this twi

8_murik_8 [283]

Answer:

4.) 7 : 4 = 36[g] : x [g]

7 * x [g] = 4 * 36 [g]

7 * x [g] = 144 [g]

x [g] = 144 [g] / 7

x [g] ≈ 20, 57 [g]

5.) 3:2 250

3x + 2x =250

5x=250

x=50

3x=3*50=150

2x=2*50=100

4 0

3 years ago

The American Community Survey showed that residents of New York City have the longest travel times to get to work compared to re

LiRa [457]

Answer:

The probability it will take a resident of the city between 17.24 and 42.13 minutes to travel to work is 0.3046.

Step-by-step explanation:

The random variable X can be defined as the travel times to get to work.

The expected travel time is, μ = 38.3 minutes.

The distribution of random variable X can be defined as the distribution of time interval between which a person reaches their work place at a constant average rate.

This implies that X follows an Exponential distribution with parameter, $\lambda=\frac{1}{\mu}=\frac{1}{38.3}$ .

The probability density function of X is:

$f_{X}(x)=\lambda e^{-\lambda x};\ x\geq 0$

Compute the probability it will take a resident of the city between 17.24 and 42.13 minutes to travel to work as follows:

$P(17.24\leq X\leq 42.13)=\int\limits^{42.13}_{17.24}{\frac{1}{38.3} e^{-x/38.3}}}\, dx$

$=\frac{1}{38.3}\times \int\limits^{42.13}_{17.24}{ e^{-x/38.3}}}\, dx$

$=\frac{1}{38.3}\times| \frac{e^{-x/38.3}}{-1/38.3}}|^{42.13}_{17.24}\\$

$=-e^{42.13/38.3}+e^{17.34/38.3}\\=-0.3329+0.6375\\=0.3046$

Thus, the probability it will take a resident of the city between 17.24 and 42.13 minutes to travel to work is 0.3046.

6 0

3 years ago