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3 years ago

Pls help this was due 3 minutes ago!!!!

Mathematics

2 answers:

ella [17]3 years ago

7 0

Answer:

-0.7

Step-by-step explanation:

2 8/10 = 2.8

2.8 + (-3.5) = -0.7

Shtirlitz [24]3 years ago

3 0

Answer:

Step-by-step explanation:

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The problem is in the pictures, please show step by step how to do it. Thanks! :)

Rama09 [41]

Answer:

-4,-5/2

Step-by-step explanation:

2x^2+3x-20 =0

2x^2+8x-5x-20 =0

2x(x+4)-5(x+4) =0

(x+4)(2x-5) =0

Either,

x+4=0

x=-4

Or,

2x-5=0

2x=5

x=5/2

7 0

3 years ago

Read 2 more answers

Using the prostate data from the faraway package with lpsa (log prostate specific antigen) as response and lcavol (log cancer vo

rewona [7]

Answer:

Step-by-step explanation:

Given that,pros.

lpsa = 1.507 + 0.719 lcavol

lpsa=0.719 lcavol + 1.507

Comparing this to equation of a straight line, y=mx+c

Then,

slope (m) = 0.719

And intercept (c) on y axis is 1.507

Then, the x axis is lcavol

Also, the y axis is lpsa.

Then the linear model preview is

Check attachment for the preview.

8 0

2 years ago

Solve by using proper methods. Show your work.

NeX [460]

Answer:

$70,201.38

Step-by-step explanation:

To determine the principal amount, we can use the formula:

$P=\dfrac{A}{(1+\frac{r}{n})^{nt}}$

A = $200,000

r = 7% or 0.07

t = 15 years

n = 12

Now let's substitute our values.

$P=\dfrac{200,000}{(1+\frac{0.07}{12})^{12(15)}}$

$P=\dfrac{200,000}{(1.00583333333)^{12(15)}}$

$P=70,201.38$

So the principal needed to get 200,000 in 15 years is $70,201.38.

7 0

2 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

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

8 0

1 year ago

Tell whether (3,5) is a solution of y = 5x – 10.

Ivanshal [37]

Answer:

yes it is

Step-by-step explanation:

plug x and y in

3 * 5 - 10 = 5 <== And this is true

6 0

2 years ago

Read 2 more answers