Ask question

Ask question

All categories

tino4ka555 [31]

2 years ago

10

One week Alaina ran 12 miles in 131.25 minutes. The next, Alaina ran 12 miles in 119.5 minutes. If she ran in a constant pace du

ring each run, about how much faster did she run each mile in he second week than in the first week?

2 answers:

vlabodo [156]2 years ago

6 0

Ou should not increase to more than 12 miles next week<span>; ... </span>25 miles<span> a </span>week<span>, usually a 7 </span>milerun<span> ... before I </span>run<span> a half. Using that method</span>

Ronch [10]2 years ago

5 0

Answer:

0.98 minutes

Step-by-step explanation:

First, we have to determine how much time it took for Alaina to run one mile on the first and the second week:

First week:

12 miles → 131.25 minutes

1 mile → x

x=(131.25 minutes*1 mile)/12 miles= 10.93 minutes

Second week:

12 miles → 119.5 minutes

1 mile → x

x= (119.5 minutes*1 mile) /12 miles= 9.95 minutes

Now, we have to find the difference between the time it took to run 1 mile in the first and the second week:

10.93-9.95= 0.98 minutes

She run 0.98 minutes faster in the second week.

You might be interested in

Find the length of the missing side. If necessary, round to the nearest tenth. side A = 38 side b =b side c = 34

ohaa [14]

A) 51, assuming this is a *right* triangle:

38^2+34^2 = 50.99^2, rounding 51!!

5 0

2 years ago

Read 2 more answers

An equation is shown below:

Dahasolnce [82]

Part A: first multiply 2 by x and -3, then combine the like terms, after that add 6 from both sides, then subtract 6x from both side, the answer is -5.
4x + 2(x-3) = 4x + 2x - 11
4x +2x -6 = 4x + 2x - 11
6x -6 = 6x - 11
6x = 6x -11 +6
6x = 6x -5
6x-6x= -5
0 = -5
part B) add the like terms

7 0

2 years ago

Read 2 more answers

What is the part and whole of 40% of 20?

pashok25 [27]

The answer is 8

I hope this helps!

5 0

2 years ago

X+2y=8 x-2y=-4 solve by system of equations

Vsevolod [243]

Answer:

Step-by-step explanation:

x + 2y = 8 -----------(I)

x - 2y = -4 -----------(II)

Add equation (I) & (II) and thus y will be eliminated and we can get the value of 'x'

(I) x + 2y = 8

(II) <u>x - 2y = -4</u><u> </u>{Now add}

2x = 4

x = 4/2

x = 2

Substitute x = 2 in equation (I)

2 + 2y = 8

2y = 8 - 2

2y = 6

y = 6/2

y = 3

3 0

2 years ago

Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.

Leni [432]

Answer:

a)

$g(2.9) \approx -6.6$

$g(3.1) \approx -3.4$

b)

The values are too small since $g''$ is positive for both values of $x$ in. I'm speaking of the $x$ values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

$y-y_1=m(x-x_1)$

where $m$ is the slope and $(x_1,y_1)$ is a point on that line.

We want to find the equation of the tangent line of the curve $g$ at the point $(3,-5)$ on $g$ .

So we know $(x_1,y_1)=(3,-5)$ .

To find $m$ , we must calculate the derivative of $g$ at $x=3$ :

$m=g'(3)=(3)^2+7=9+7=16$ .

So the equation of the tangent line to curve $g$ at $(3,-5)$ is:

$y-(-5)=16(x-3)$ .

I'm going to solve this for $y$ .

$y-(-5)=16(x-3)$

$y+5=16(x-3)$

Subtract 5 on both sides:

$y=16(x-3)-5$

What this means is for values $x$ near $x=3$ is that:

$g(x) \approx 16(x-3)-5$ .

Let's evaluate this approximation function for $g(2.9)$ .

$g(2.9) \approx 16(2.9-3)-5$

$g(2.9) \approx 16(-.1)-5$

$g(2.9) \approx -1.6-5$

$g(2.9) \approx -6.6$

Let's evaluate this approximation function for $g(3.1)$ .

$g(3.1) \approx 16(3.1-3)-5$

$g(3.1) \approx 16(.1)-5$

$g(3.1) \approx 1.6-5$

$g(3.1) \approx -3.4$

b) To determine if these are over approximations or under approximations I will require the second derivative.

If $g''$ is positive, then it leads to underestimation (since the curve is concave up at that number).

If $g''$ is negative, then it leads to overestimation (since the curve is concave down at that number).

$g'(x)=x^2+7$

$g''(x)=2x+0$

$g''(x)=2x$

$2x$ is positive for $x>0$ .

$2x$ is negative for $x$ .

That is, $g''(2.9)>0 \text{ and } g''(3.1)>0$ .

So $2x$ is positive for both values of $x$ which means that the values we found in part (a) are underestimations.

6 0

3 years ago