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2 years ago

Chloe drives from her house to her Grandma's house 48 miles away in an hour and 12 minutes. How fast was she driving on average?

2 answers:

5 0

Working in minutes might give us a little more flexibility here, so we can start by converting 1 hour and 12 minutes, giving us 60 + 12 = 72 minutes for Chloe to drive 48 miles.

We can write her Chloe's rate then as 48 mi/72 min; since 48 and 72 have the factor 12 in common, we can reduce this to a rate of 4 mi/6 min. We want to find how many miles Chloe would travel in <em>60 minutes </em>at that rate. Since 60 = 6 x 10, we can scale the rate up by a factor of 10, getting us a rate of 4 x 10 = 40 mi/60 min, or 40 mph.

Lera25 [3.4K]2 years ago

3 0

To find a speed in miles per hour, we need to divide a number of miles by a number of hours.

We are given 48 miles, so we already have a number of miles.

We are given the time as 1 hour and 12 minutes. We need to convert this time into hours.

1 hour + 12 minutes = 1 hour + 12/60 hour = 1.2 hours

speed = distance/time = 48 miles / 1.2 hours = 40 mph

Answer: The fourth choice, 40 mph.

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1 year ago

Read 2 more answers

Solve the system by substitution.<br> -2x + 7y = 15<br> бу = 3

Salsk061 [2.6K]

Answer:

Step-by-step explanation:

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2 years ago

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$