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2 years ago

9

For f(x) = 4x+1 and g(x) = x^2 -5, find (f-g)(x)

1 answer:

d1i1m1o1n [39]2 years ago

4 0

Answer:

(f-g)(x) is given by -x^2+4x+6

Step-by-step explanation:

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The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initi

slega [8]

Answer:

The correct option is 4

Step-by-step explanation:

The solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}$

Now for the initial condition the value of C is calculated as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16$

So the solution is given as

$y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}$

Simplifying the equation as

$y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}$

So the correct option is 4

8 0

3 years ago

Read 2 more answers

Write an equation of each line in standard form with integer coefficients.<br> 4. y = 2.4x + 1.8

Nadusha1986 [10]

Answer:

<h2>12x - 20y = -9</h2>

Step-by-step explanation:

The standard form of an equation of a line:

$Ax+By=C$

We have

$4y=2.4x+1.8$

Convert

$4y=2.4x+1.8$ <em>multiply both sides by 10</em>

$40y=24x+18$ <em>subtract 24x from both sides</em>

$-24x+40y=18$ <em>change the signs</em>

$24x-40y=-18$ <em>divide both sides by 2</em>

$12x-20y=-9$

3 0

3 years ago

Which figure can be formed from the net?

Law Incorporation [45]

Answer:

#1 is the answere

Step-by-step explanation:

Check the sides

6 0

2 years ago

The distance between City A and City B is 200 miles. A length of 2.1 feet represents this distance on a certain wall map. City C

ratelena [41]

Answer:

300 miles

Step-by-step explanation:

200/2.1= 95.238...

95.238...*3.15=300

Hope this helps :)

3 0

3 years ago

Read 2 more answers

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

2 years ago