Ask question

Ask question

All categories

2 years ago

14

Rack runners will all cover the same distance. The winner will take: Question 2 options: more time than the other runners less t

ime than the other runners the same time as the other runners half as much time as the other runners

1 answer:

Damm [24]2 years ago

6 0

Answer:

yup yup yup yup yup yup yup yup yup

You might be interested in

What times what equals -20 and added together equals -8

Leto [7]

4 x -5 = -20 then 12 + -20 = -8

5 0

3 years ago

In a parallelogram, opposite angles and opposite sides have the same measure. The sides of a parallelogram are 6 cm and 4 cm lon

Illusion [34]

Answer:

i dont know

Step-by-step explanation:

i dont know

8 0

3 years ago

What is the sum of the distances of R to Q and P to Q?

Shalnov [3]

The sum of the distances of R to Q and P to Q is 9 units.

Solution:

Given points are P(–3, 6), Q(3, 6) and R(3, 3).

Distance between two points formula:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance from R to Q:

Here $x_1=3, y_1=3, x_2=3, y_2=6$

Substitute these in the given formula, we get

Distance = $\sqrt{(3-3)^2+(6-3)^2}$

$=\sqrt{0+9}$

= 3 units

Distance from R to Q is 3 units.

Distance from P to Q:

Here $x_1=-3, y_1=6, x_2=3, y_2=6$

Substitute these in the given formula, we get

Distance = $\sqrt{(3-(-3))^2+(6-6)^2}$

$=\sqrt{36+0}$

= 6 units

Distance from P to Q is 6 units.

Sum of the distances = 3 + 6 = 9 units

Hence the sum of the distances of R to Q and P to Q is 9 units.

4 0

3 years ago

Area=<br> Help me please!! Thanks so much-needed

Anettt [7]

First you have to figure out the side length of the triangle.

sin(60)= X/8

Let's call x the side length.

8 sin(60)=X

X=6.93

The multiply it by 4

The area is 27.72

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cleft%20%28%20%5Cbigg%28%20%5Cint_%7B1%7D%5E%7B%20%7B

Margaret [11]

Applying the product rule gives

$\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}$

Use the fundamental theorem of calculus to compute the remaining derivatives.

$\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt$

The remaining integrals are

$\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}$

$\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)$

and so the overall derivative is

$\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)$

which could be simplified further.

5 0

2 years ago