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2 years ago

8

Explain constraint of domain on parabola graphs

1 answer:

scZoUnD [109]2 years ago

6 0

Answer:The set of possible values for the independent variable is called the domain. Similarly, there are constraints on the values for any dependent variable. ... Graphs of functions are traditionally drawn with the independent variable on the horizontal axis and the dependent variable on the vertical axis.

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago

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nirvana33 [79]

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3 years ago

How do ypu find the area of 1/4 of a circle?

julia-pushkina [17]

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2 years ago

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

3 years ago

betty closes the nozzle and fills it completely with a liquid. She then opens the nozzle. If the liquid drips at the rate of 14

natta225 [31]

Answer:

4.71 minutes

Step-by-step explanation:

Incomplete question [See comment for complete question]

Given

Shape: Cone

$r = 3$ -- radius

$h = 7$ --- height

$Rate = 14in^3/min$

Required

Time to pass out all liquid

First, calculate the volume of the cone.

This is calculated as:

$V = \frac{1}{3} \pi r^2h$

This gives:

$V = \frac{1}{3} * 3.14 * 3^2 * 7$

$V = \frac{1}{3} * 197.82$

$V = 65.94in^3$

To calculate the time, we make use of the following rate formula.

$Rate = \frac{Volume}{Time}$

Make Time the subject

$Time= \frac{Volume}{Rate }$

This gives:

$Time= \frac{65.94in^3}{14in^3/min}$

$Time= \frac{65.94in^3}{14in^3}min$

Cancel out the units

$Time= \frac{65.94}{14} min$

$Time= 4.71 min\\$

3 0

2 years ago