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2 years ago

Trong mặt phẳng tọa độ, viết chương trình đường tròn (C') là ảnh của (C): x^2 +y^2 -2x+4y=0

Mathematics

1 answer:

ryzh [129]2 years ago

3 0

Step-by-step explanation:

Find the properties of the conic section

−

x2+y2-2x+4y.0

Center:

(1-2)

Radius:

√5

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What is an equation of the line that passes through the points (4,-1) and (8,5)?

fiasKO [112]

Answer:

y=3/2x-7

Step-by-step explanation:

the equation of the line for slope-intercept form is y=mx+b, where m is the slope and b is the y intercept.

we are given two points: (4,-1) and (8,5)

the equation for slope is (y2-y1)/(x2-x1)

label the points:

x1=4

y1=-1

x2=8

y2=5

now substitute into the equation:

m=(5--1)/(8-4)

m=6/4

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here is our equation so far:

y=3/2x+b

we need to find b

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3 years ago

What function is represented in the table?

Pani-rosa [81]

Answer:

Step-by-step explanation:

an exponential function of the form

y = a $b^{x}$

to find a and b use ordered pairs from the table

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0.5 = a $b^{0}$ [ $b^{0}$ = 1 ] , then

a = 0.5

so y = 0.5 $b^{x}$

using (1, 2 ) , then

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2 years ago

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Alenkasestr [34]

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I think this right : )

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3 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

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3 years ago