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3 years ago

Y=(1/2)^x +2. What is the horizontal asymptote of this function?

1 answer:

6 0

$\lim_{x\to -\infty} \left(\frac{1}{2}\right)^x+2=\infty+2=\infty\\
\lim_{x\to \infty} \left(\frac{1}{2}\right)^x+2=0+2=2\\$

There is one-sided horizontal asymptote $y=2$

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One serving of Gooey Gushers provides 12% of the daily value of vitamin C. Tim ate 312 servings of Gooey Gushers.

Studentka2010 [4]

Answer:

a) Given expression,

12% of 312

Since 100% of 312 = 312

⇒ 10% of 312 = 31.2,

⇒ 1% = 3.12

⇒ 2% = 2 × 3.12 = 6.24

So, 12% of 312 = 10% of 312 + 2% of 312

= 31.2 + 6.24

= 37.44

b) Given,

Total number of servings = 312,

The percentage of vitamin C in each serving = 12%,

Thus, the total amount of the vitamin C in 312 servings = 12% of 312

$=\frac{12\times 312}{100}$

$=\frac{3744}{100}$

= 37.44,

Hence, the question can be asked about the total amount of the vitamin C.

6 0

3 years ago

Jenny bought a 5 notebooks and 2 pencils, and the total was 9 dollars.

weeeeeb [17]

You would solve this with simultaneous equations, so if we write it as:
5n + 2p = 9
3n + 2p = 6
(subtract)
2n = 3
÷ 2
notebooks = 1.5

Now you would substitute it in:
(3 × 1.5) + 2p = 6
4.5 + 2p = 6
- 4.5
2p = 1.5
÷ 2
pens = 0.75

So your final answer is notebooks are $1.50 and pens are $0.75, I hope this helps!

6 0

3 years ago

30 points! You start at (-2, 2). You move left 3 units and right 10 units. Where do you end?

Cerrena [4.2K]

Answer:

The answer is (5, 2)

Step-by-step explanation:

I. If move left = -x

move right = +x

So -2 - 3 = -5 => move left

-5 + 10 = 5 => move right

I suggest that move left and right is x cordinate

The answer is (5,2)

Is that correct?

7 0

3 years ago

Which is the solution to the following system? y=3x+1 6x-2y=4

zalisa [80]

Answer:

x=-7/3 and y=-6

Step-by-step explanation:

6x-3(3x+1)=4

6x-9x-3=4

-3x-3=4

-3x=7

x=-7/3

y=3(-7/3)+1

y=-6

8 0

3 years ago

Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4

OLEGan [10]

Answer:

$y(0.2)=3$ , $y(0.4)=3.005974448$ , $y(0.6)=3.017852169$ , $y(0.8)=3.035458382$ , and $y(1.0)=3.058523645$

The general solution is $y=\ln \left(\frac{3t^2}{2}+e^3\right)$

The error in the approximations to y(0.2), y(0.6), and y(1):

$|y(0.2)-y_{1}|=0.002982771$

$|y(0.6)-y_{3}|=0.008677796$

$|y(1)-y_{5}|=0.013499859$

Step-by-step explanation:

Point a:

The Euler's method states that:

$y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right)$ where $t_{n+1}=t_n + h$

We have that $h=0.2$ , $t_{0}=0$ , $y_{0} =3$ , $f(t,y)=3te^{-y}$

We need to find $y(0.2)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{1}=t_{0}+h=0+0.2=0.2$

$y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=$

$=3 + 0.2 \cdot \left(0 \right)= 3$

$y(0.2)=3$

We need to find $y(0.4)$ for $y'=3te^{-y}$ , when $y(0)=3$ , $h=0.2$ using the Euler's method.

So you need to:

$t_{2}=t_{1}+h=0.2+0.2=0.4$

$y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=$

$=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448$

$y(0.4)=3.005974448$

The Euler's Method is detailed in the following table.

Point b:

To find the general solution of $y'=3te^{-y}$ you need to:

Rewrite in the form of a first order separable ODE:

$e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt$

Integrate each side:

$\int \:e^ydy=e^y+C$

$\int \:3t\:dt=\frac{3t^2}{2}+C$

$e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}$

We know the initial condition y(0) = 3, we are going to use it to find the value of $C_{1}$

$e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3$

So we have:

$e^y=\frac{3t^2}{2}+e^3$

Solving for y we get:

$\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)$