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3 years ago

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The distance from Kenya's house to her job is 18 miles. One morning she left her house at 7:15 a.m. She traveled at the rate of

45 miles per hour for 20 minutes, but to traffic she had to slow down to 20 miles per hour. At what time did Kenya arrive at work?

1 answer:

Snezhnost [94]3 years ago

8 0

20 minutes = 1/3 of an hour

1/3 = 0.333

1/3 x 45mph = 15, so she drove 15 miles at 45mph

18-15 = 3

3 miles/ 20 mph = 0.15 hours

0.333+0.15 = 0.483 hours total

0.483 x 60 = 28.98 so approximately 29 minutes total driving time

7:15 + 29 minutes is 7:44 am she arrived at work

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PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

----------------------------------------

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

--------------------------

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

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