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2 years ago

15

The exponential function f(x) has a horizontal asymptote at y = 3. What is the

1 answer:

fenix001 [56]2 years ago

3 0

Answer:

The function diverges to $+\infty$ .

Step-by-step explanation:

The function diverges to $+\infty$ .

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OI<br> I<br> NEED<br> HELP<br> PLOZ<br> COME<br> OVER<br> HERE<br> AND<br> HELP<br> ME

bija089 [108]

The answer is C, let me know if i am correct!!

6 0

2 years ago

Read 2 more answers

What percent of 641 is 73?

alexira [117]

You can divide 73/641 to find the percent:

73/641 = 0.11388455...

It keeps going on so I'll stop there.

Now to make it into a percentage, multiple it by 100:

0.11388455 * 100 = 11.388455%

3 0

3 years ago

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

2 years ago

Two sides of an isosceles triangle measure 3 inches and 7 inches. Which could be the length of the third side?

slamgirl [31]

Answer:

7

Step-by-step explanation:

6 0

2 years ago

13 - 8m = 21<br> por favor ayuda

Ulleksa [173]

Si estás tratando a encontrar M este es
13-8m=21
8m=8
M=1

6 0

2 years ago