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2 years ago

Write an equation for three-fourth of m is 20

Mathematics

2 answers:

alisha [4.7K]2 years ago

4 0

Answer:

Since, three fourth of $t$ is $15$. Hence, the required equation is $\frac{3t}{4}=15$

Step-by-step explanation:

Oksanka [162]2 years ago

3 0

Answer:

$= > \frac{3}{4} \times m = 20$

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You can walk 1 1/2 miles in 1/2 hour what is your average speed

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Let’s turn the mixed number into a fraction
3/2 x 1/2
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3/4
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3 years ago

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What is Coordinate rotation

saul85 [17]

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Replace x with y and vice versa

Step-by-step explanation:

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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

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

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

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

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

with $0\le t\le1$ . Then

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

and

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

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1 year ago

A new cookie cake company is offering a

gtnhenbr [62]

Answer:

200.96in^2

Step-by-step explanation:

If the 1st cookie cake is 8" for its diameter and the 2nd cookie cake will have twice the length diameter of the 1st cookie cake .You will do 2 × 8 to get the diameter of the 2nd cookie cake.

First cookie cake diameter is 8".

Second cookie cake's diameter is 16".

Since we need the radius to find the area of the 2nd cookie cake we're going to do 16 / 2 which equals 8 so 8 is the radius.

A=(pi)r^2

A=(pi)8^2 *to the 2nd power means times by the same number

A=(pi)64

A= (3.14)(64)

A=200.96

=200.96^2

pi= 3.14 *for this problem

7 0

3 years ago

Wen takes a job with a starting salary of $100,000 for the first year. she earns a 3% increase each year. what does s5 represent

steposvetlana [31]

$S_{n} = \frac{ a_{1}(1- r^{n}) }{1-r}$ $a_{1} = 100000, r=1.03, n=5$
$S_{5} = \frac{100000(1- 1.03^{5}) }{1-1.03} = \frac{100000(-.159)}{-.03} =530,000$

So after 5 years she is making $530,000.

Hope that helps

8 0

3 years ago

Read 2 more answers

Write an equation for three-fourth of m is 20​

Write an equation for three-fourth of m is 20