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kakasveta [241]
2 years ago
15

A truck has left Amsterdam with a speed of 52 mph. After 5 hours a car has left Amsterdam with a speed of 78 mph. At what distan

ce from Amsterdam will the car catch up with the truck? Answer should be in miles.
Mathematics
1 answer:
UkoKoshka [18]2 years ago
8 0

Answer:360 miles


Step-by-step explanation:


You might be interested in
The means and mean absolute deviations of the individual times of members on two 4 times 400 meter relay track teams are shown i
Yakvenalex [24]

The correct answer is option C which is the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B will be 0.25.

<h3>What is mean?</h3>

Mean is defined as the ratio of the sum of the number of the data sets to the total number of the data.

The difference between the mean times is about 16 times the mean absolute deviation of the data set.

59.32 - 59.1 = 0.22

2.4 - 1.5 = 0.9

0.22 / 0.9 = 0.25

Therefore the correct answer is option C which is the ratio of the difference in the means of the two teams to the mean absolute deviation of Team B will be 0.25.

To know more about mean follow

brainly.com/question/968894

#SPJ1

7 0
1 year ago
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

7 0
3 years ago
Read 2 more answers
1. What does it mean for a number to have a variable at the end of it? (i.e 1.6x)
miskamm [114]

Answer:

1. 1.6x has a variable which means an unknown number

2. 5 more than 0.10 times a number

Step-by-step explanation:

A variable is an unknown number represented by a letter. It provides a way to write a relationship in the real world without knowing all the values involved, 1.6x means 1.6 times an unknown number mathematically. A real world example would be a percent growth of 60% on an unknown price - 1.6x.

5 + 0.10x means 5 more than 0.10 times an unknown number. A real world example here would be "10% off a sale plus $5 coupon". We don't know the price of the item but we do know it is 10% off and we get $5 off. Once we know the price we input it int for x and find the total discount.

8 0
3 years ago
What’s the equation for a line passing through points (-1, 4.5) and (6, 8)?
san4es73 [151]

Answer:

<em>Given </em><em>points </em><em>(</em><em> </em><em>-</em><em>1</em><em> </em><em>,</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>)</em><em> </em><em>and </em><em>(</em><em> </em><em>6</em><em> </em><em>,</em><em> </em><em>8</em><em> </em><em>)</em>

Step-by-step explanation:

<em>first </em><em>we </em><em>should </em><em>calculate </em><em>slope </em>

<em>Slope </em><em>(</em><em> </em><em>m</em><em>) </em>

<em>=</em><em> </em><em>(</em><em> </em><em>y2 </em><em>-</em><em> </em><em>y1</em><em>) </em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>x2</em><em> </em><em>-</em><em> </em><em>x1</em><em>) </em>

<em>=</em><em> </em><em>(</em><em> </em><em>8</em><em> </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>)</em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>6</em><em> </em><em> </em><em>+</em><em> </em><em>1</em><em> </em><em>)</em>

<em>=</em><em> </em><em>3</em><em>.</em><em>5</em><em>/</em><em>7</em>

<em>=</em><em> </em><em>0</em><em>.</em><em>5</em>

<em>Now </em><em>the </em><em>equation </em><em>of </em><em>line </em><em>is </em><em>given </em><em>by </em>

<em>y </em><em>-</em><em> </em><em>y1 </em><em>=</em><em> </em><em>m </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>x</em><em>1</em><em> </em><em>)</em>

<em>y </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>(</em><em> </em><em>x </em><em>+</em><em> </em><em>1</em><em> </em><em>)</em>

<em>(</em><em> </em><em>y </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em> </em><em>)</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>(</em><em> </em><em>x </em><em>+</em><em> </em><em>1</em><em>)</em>

<em>y </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>0</em><em>.</em><em>5</em>

<em>y </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>+</em><em> </em><em>4</em><em>.</em><em>5</em>

<em>y </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>5</em>

3 0
2 years ago
Create a unique parabola in the pattern f(x) = (x − a)(x − b)
MissTica

Answer:

f(x)=(x-6)(x-(-7))

Step-by-step explanation:

Foiled:

0 = x + x - 42

5 0
2 years ago
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