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Naddika [18.5K]

3 years ago

8

Item 19 A plant has an initial height of 1 inch and grows at a constant rate of 3 inches each month. A second plant that also gr

ows at a constant rate has an initial height of 4 inches and is 28 inches tall after 1 year. After how many months are the plants the same height?

1 answer:

Alex777 [14]3 years ago

3 0

3 months, i just answered this same question lol.

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What is the range of the function on the graph?

Marta_Voda [28]

Answer:

no. 4

Step-by-step explanation:

the numbers start at two and keeps going up.

3 0

2 years ago

Read 2 more answers

An arc is 6.5 cm long and it subtends a central angle of 45 degrees. Find the radius of the circle.

Orlov [11]

Arc length = radius * central angle (measured in radians)
There are <span> <span> <span> 57.2957795131 </span> </span> </span> degrees per radian so
45 degrees = (45 / <span> <span> <span> 57.2957795131) = </span></span></span> <span> <span> <span> 0.7853981634 </span> </span> </span> radians<span><span> </span> </span>
radius = arc length / central angle (radians)
radius = 6.5 / <span> <span> 0.7853981634 </span> </span> radians = <span> <span> <span> 8.2760570408 </span> </span> </span> cm

http://www.1728.org/radians.htm

8 0

2 years ago

a random sample size of 175 is drawn from a large population. The population standard deviation is 13.6. The sample mean is 87.3

navik [9.2K]

Answer:

The size of the population mean would be provided as:

mean +/- std/sqrt(sample size)

<=>

87.3 +/- 13.6/sqrt(175)

<=>

87.3 +/- 1.03

=> Option B is correct.

Hope this helps!

:)

6 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

What is the solution for the following system of equations?<br><br>5x + 7y = 3<br><br>2x + 3y = 1

spayn [35]

<span>A) 5x + 7y = 3
B) 2x + 3y = 1

Multiplying Equation B by -2.5
</span><span>B) -5x -7.5y = -2.5 Then adding this to Equation A)
A) 5x + 7y = 3</span>

-.5y = .5
y = -1

Since
<span>2x + 3y = 1 then
2x -3 = 1
then x = 2

</span>

6 0

3 years ago