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2 years ago

How many minutes per lap does it take a runner who after 45 minutes has run 9 laps?

Mathematics

1 answer:

trapecia [35]2 years ago

3 0

Answer:

5 minutes per lap.

Step-by-step explanation:

45÷9=5

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AleksAgata [21]

Answer: 3 • ? = -6

Reasoning: -6/ 3 = -2
3 • -2 = -6

8 0

2 years ago

How is simplifying similar to division?

Alex

When you simpilfy a fraction you make it smaller. When you divide a number you make it smaller.

Ex: 5 /5 1
_ = _
10 /5 2

3 0

3 years ago

Read 2 more answers

What is the axis of symmetry

Verdich [7]

Answer:

its is x=0

Step-by-step explanation:

5 0

3 years ago

EXPLAIN THE ERROR Cody was given two points (1/2,4) and (2/3,1), on a line and asked to create a linear equation in standard for

inessss [21]

Answer:

Cody made a mistake when calculating the slope of the line and this affected every other steps after than.

She divided -3 by 6 instead of -3 by 1/6

Step-by-step explanation:

Given

$(x_1,y_1) = (\frac{1}{2},4)$

$(x_2,y_2) = (\frac{2}{3},1)$

<em>See attached for steps</em>

Required

Explain Cody's error

<em>Cody made a mistake when calculating the slope of the line and this affected every other steps after than.</em>

See Proof

Slope (m) is calculated as thus:

$m =\frac{y_2 - y_1}{x_2 - x_1}$

$m =\frac{1 - 4}{\frac{2}{3} - \frac{1}{2}}$

$m =-\frac{ 3}{\frac{1}{6}}$

$m = -3/\frac{1}{6}$

$m = -3 * 6$

$m = -18$

This is in contrast to $m = -\frac{1}{2}$ , calculated by Cody

Solving further to determine the equation.

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

Where

$(x_1,y_1) = (\frac{1}{2},4)$

$m = -18$

$y - 4 = -18(x - \frac{1}{2})$

$y - 4 = -18x + 9$

Collect Like Terms

$y + 18x = 9 + 4$

$y + 18x = 13$

6 0

3 years ago

Determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist

yuradex [85]

Answer:

0 < t < 5 is the required interval for the differential equation (t - 5)y' + (ln t)y = 6t to have a solution.

Step-by-step explanation:

Given the differential equation

(t - 5)y' + (ln t)y = 6t

and the condition y(1) = 6

We can rewrite the differential equation by dividing it by (t - 5) as

y' + [(ln t)/(t - 5)]y = 6t/(t - 5)

(ln t)/(t - 5) is continuous on the interval (0, 5) and (5, +infinity).

6t/(t - 5) is continuous on (-infinity, 5) and (5, +infinity)

We see that for these expressions, we have continuity at the intervals (0, 5) and (5, +infinity).

But the initial condition is y = 6, when t = 1.

The solution to differential equation is certain to exist at (0, 5)

Which implies that

0 < t < 5

is the required interval.

3 0

3 years ago