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Marysya12 [62]
3 years ago
11

on another training ride jana bikes 15 miles in 50 minutes explain how you could find the number of miles she bikes in 1 hour ex

plain please
Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
5 0
So she is going 15 miles every 50 minutes...
arsen [322]3 years ago
3 0
Hiii I think that she wrode her bike 18 miles in one hour because 3 times 5 equals 15 miles and 15 miles equals 50 minutes. So if you add another 3 miles to 15 you get 18 miles in one hour
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Only answer #5 <br><br><br>Please help me and thx and if u can answer #6 too if u can
Marianna [84]

Answer:

5. D

6. C

Step-by-step explanation:

<u>Question 5</u>

Lets plug in values of x and y into each equation to see which one works for the entire data set...

y=2x-4

1) Plug in -7 as x and -18 as y:

-18=2(-7)-4

2) Multiply:

-18=-14-4

3) Subtract:

-18=-18

The answers are the same, therefore the correct answer to question 5 is equation D. Keep in mind, you want to check ever single pair of points to make sure it works for all of them, I'm just not going to list that as it would make this answer a lot longer for no reason.

<u>Question 6:</u>

We know that the cost for the boat is $25 as a base price, and also costs an additional $1.30 per day for fuel.

The correct equation is C=25+1.30d, because we are adding 1.3 every single day.

4 0
3 years ago
Calculus 3 help please.​
Reptile [31]

I assume each path C is oriented positively/counterclockwise.

(a) Parameterize C by

\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}

with -\frac\pi2\le t\le\frac\pi2. Then the line element is

ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt

and the integral reduces to

\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt

The integrand is symmetric about t=0, so

\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt

Substitute u=\sin(t) and du=\cos(t)\,dt. Then we get

\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}

(b) Parameterize C by

\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{3^2+4^2} \, dt = 5\,dt

and

\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt

Integrate by parts with

u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}

\displaystyle \int u\,dv = uv - \int v\,du

\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}

(c) Parameterize C by

\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt

and

\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}

8 0
1 year ago
Después de 15 meses de haber prestado un capital al 5% de rédito, tengo que pagar de interés Q468.75. ¿Qué capital se ha prestad
Aleonysh [2.5K]

The capital borrowed, or the principal when the interest was Q468.75 is <u>Q7500</u>.

The principal P, amount borrowed, at a certain rate of interest R%, for  a time of T years, giving an interest of I, can be calculated using the formula:

P = (I * 100)/(R * T).

In the question, we are asked to find the capital borrowed, that is, the principal, when the user pays Q468.75 after 15 months at 5% of income.

Thus, Interest (I) = Q468.75, rate of interest (R) = 5%, and time (T) = 15 months = 15/12 years = 1.25 years.

Thus, the principal P, can be calculated by substituting the values in the formula: P = (I * 100)/(R * T).

P = (468.75*100)/(5*1.25),

or, P = 46875/6.25,

or, P = 7500.

Thus, the capital borrowed, or the principal when the interest was Q468.75 is <u>Q7500</u>.

The provided question is in Spanish. The question in English is:

"After 15 months of having borrowed capital at 5% of income, I have to pay Q468.75 in interest. What capital has been borrowed?"

Learn more about Interest at

brainly.com/question/25793394

#SPJ1

5 0
2 years ago
Change the following quantity into the indicated unit.
mafiozo [28]

Answer:

11.8 in.

Step-by-step explanation:

8 0
4 years ago
Please answer!!!! please!!!!!!
olga55 [171]

The problem here is:  Too many words.

If you boil the problem down to the bare facts, the question is:

             "$210 is 15% of what number ?"

I'm pretty sure you can take it from there,
but here's the rest of it anyway:

                                                $210  =  15% of (March amount)

                                                $210  =  0.15 x (March amount)

Divide each side by  0.15 :     $1400 = March amount .

Check:
1400 + 210 =  1610
1.15 x 1400 = 1610                                      yay!
5 0
3 years ago
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