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3 years ago

14

Kamani can run 1 lap in 6 minutes. How many laps can she run in 18 minutes? If she runs 7 laps every day, how many minutes does

she run everyday?

1 answer:

cricket20 [7]3 years ago

4 0

3 laps in 18 minutes & 42 minutes a day

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Quiana has 5 times more text messages from Yolanda than from Pierre. Quiana has 30 text messages from Yolanda. Which equation ca

VladimirAG [237]

Answer:

$p=\frac{30}{5}$

Step-by-step explanation:

Quiana has 5 times more text messages from Yolanda than from Pierre, we also know that she actually has 30 text messages from Yolanda. In other words, 30 texts from Yolanda is 5 times more text messages than from Pierre.

Thus, if we call p the number of texts messages from Pierre we would have that

$p=\frac{30}{5}$ (since 30 texts from Yolanda is 5 times more than p, we can divide by 5 to find p)

And if we solve this equation we would get that Quiana has 6 text messages from Pierre

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3 years ago

Is diameter and radius the same thing

fredd [130]

radius* 2=diameter so no diameter and radius aren't the same thing because radius is half the distance across the circle and diameter is the full distance across the circle.

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3 years ago

Read 2 more answers

Which side lengths form a right triangle?

saw5 [17]

B
because
${1}^{2} + {8}^{2} = {( \sqrt{65}) }^{2}$

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3 years ago

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What information can be used to compare linear relationships? Explain why.

stellarik [79]

Answer:the slope and the y intercept (you can tell there linear if the slope is the same and the y-intercepts are different).

Step-by-step explanation:

8 0

3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago