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irina1246 [14]
3 years ago
12

Rose can run 4 miles in 56 minutes how many miles does Rosa run if she runs for 42 minutes

Mathematics
2 answers:
sdas [7]3 years ago
5 0

Answer:

<em>At the same rate Rose will run</em><em> 3 miles</em><em> in 42 minutes.</em>

Step-by-step explanation:

Hint- This problem ca be solved using Unitary method.

This Here given that, Rose can run 4 miles in 56 minutes.

i.e Rose in 56 minutes can run 4 miles.

So Rose in 1 minutes can run \dfrac{4}{56} miles.

So Rose in 42 minutes can run \dfrac{4}{56}\times 42=3 miles.

Therefore, if Rose can run 4 miles in 56 minutes, she will run 3 miles in 42 minutes.

kogti [31]3 years ago
4 0
\frac{56}{42} = \frac{4}{x}  then you figure out what you divide 56 by to get 4. which is 14. then you divide 42 by the same number 14.

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Landon is going to invest in an account paying an interest rate of 4.3% compounded continuously. How much would Landon need to i
Mazyrski [523]

Answer:

$ 6,189.18

Step-by-step explanation:

From the above question, we can deduce that we are meant to find the Principal (Initial Amount ) invested.

The formula for the Principal of a compound interest that is compounded continuously is given as:

P = A / e^rt

Where

P = Principal

A = Totally Amount after time t = $11,300

r = Interest rate = 4.3 % = 0.043

t = 14 years

P = $11,300/ e ^0.043 × 14

P = $ 6,189.18

Hence, Landon needs to invest, $ 6,189.18

5 0
3 years ago
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Charra [1.4K]
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7 0
3 years ago
What is the perimeter of a quadrilateral with vertices at (3,-6),(8,-6),(3,-4), and (8,-4)
Eduardwww [97]

Answer:

im sorry i dont know but this is for a challenge

Step-by-step explanation:

3 0
2 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
Find the length of the curve. R(t) = 7t, 3 cos(t), 3 sin(t) , −2 ≤ t ≤ 2
Leviafan [203]

we are given

r(t)=(7t,3cos(t),3sin(t))

we can find x , y and z

x=7t,y=3cos(t),z=3sin(t))

now, we can use arc length formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

now, we can find derivative

x'=7,y'=-3sin(t),z'=3cos(t))

now, we can plug values

and we get

L=\int _{-2}^2\sqrt{7^2+\left(-3\sin \left(t\right)\right)^2+\left(3\cos \left(t\right)\right)^2}dt

L=\int _{-2}^2\sqrt{7^2+9}dt

=\sqrt{58}\cdot \:2-\left(-2\sqrt{58}\right)

L=4\sqrt{58}...........Answer


8 0
3 years ago
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