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2 years ago

Centro en (4,-1) y pasa por el punto (-1,3)

Mathematics

1 answer:

maks197457 [2]2 years ago

7 0

Answer:

yniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehcsacasvavadvadvsdvsdvsdvdvdvyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh ollehyniar hcum oot sti ereh pleh esaelp retaw eht rednu ma i uoy era woh olleh

Step-by-step explanation:

You might be interested in

Can anyone give me any answers for the first 2 in each coloured section please

blsea [12.9K]

A.) x=5

B.) x=4

C.) x=8

D.) x=2

E.) x=7

F.) x=6

G.) x=2

H.) x= -3

I.) x=5

J.) x=8

K.) x=9

L.) x=5

M.) fraction form; x=36/7 decimal form; x=5.142857 Mixed number form; x= 5 1/7

N.) x=7

Hope this helps

6 0

3 years ago

Write the equation of the blue line using slope intercept form (y=mx+b)

dalvyx [7]

Answer:

Step-by-step explanation:

X1=-4

Y1=7

X2=-3

Y2=6

(-2,0 ) X INTERCEPT

(0,-7)Y INTERCEPT OR B

M=6-7/-3+4

M=_1

Y=-1X+-7

5 0

2 years ago

Nick randomly selects a digit from the set { 0 , 1 , 2 , . . . , 9 } and a letter from the set { A, B, C, . . ., Z } . Matthew w

Alex787 [66]

Answer: 259/260 or 0.99615 (depending on which answer format your teacher wants)

There are 10 numbers in the set {0, 1, 2, ..., 9}. There are 26 letters in the set {A, B, C, ..., Z}. Multiply those values: 10*26 = 260. So there are 260 ways to pick a number followed by a letter. One example is 7P.

There is only one way Matthew can get the correct answer, and there are 260 - 1 = 259 ways to get the wrong answer. We divide 259 over 260 to get the probability of getting the incorrect answer, which is 259/260.

If you need this fraction in decimal form, then use a calculator to find that 259/260 = 0.99615 approximately

7 0

3 years ago

Given f(x) = 4x^4 find f^-1(x) Then state whether f^-1(x) is a function.

sdas [7]

Answer:

$f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$
$f^{-1}(x) \quad\text{is not a function}$

Step-by-step explanation:

To find the inverse function, solve for y:

$x=f(y)\\\\x=4y^4\\\\\dfrac{x}{4}=y^4\\\\\pm\sqrt[4]{\dfrac{x}{4}}=y\\\\f^{-1}(x)=\pm\sqrt[4]{\dfrac{x}{4}}$

f(x) is an even function, so f(-x) = f(x). Then the inverse relation is double-valued: for any given y, there can be either of two x-values that will give that result.

___

A function is single-valued. That means any given domain value maps to exactly one range value. The test of this is the "vertical line test." If a vertical line intersects the graph in more than one point, then that x-value maps to more than one y-value.

The horizontal line test is similar. It is used to determine whether a function has an inverse function. If a horizontal line intersects the graph in more than one place, the inverse relation is not a function.

Since the inverse relation for the given f(x) maps every x to two y-values, it is not a function. You can also tell this by the fact that f(x) is an even function, so does not pass the horizontal line test. When f(x) doesn't pass the horizontal line test, f^-1(x) cannot pass the vertical line test.

_____

The attached graph shows the inverse relation (called f₁(x)). It also shows a vertical line intersecting that graph in more than one place.

5 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago