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NemiM [27]
3 years ago
14

Find the value of x for the following equivalent functions: f(x)=2x^2−3x+4 and f(x)=2x^2−1

Mathematics
1 answer:
masha68 [24]3 years ago
3 0
A) y = 2x – 7  and f(x) = 7 – 2xIncorrect. These equations look similar but are not the same. The first has a slope of 2 and a y-intercept of −7. The second function has a slope of −2 and a y-intercept of 7. It slopes in the opposite direction. They do not produce the same graph, so they are not the same function. The correct answer is f(x) = 3x2 + 5 and y = 3x2 + 5. B) 3x = y – 2  and f(x) = 3x – 2Incorrect. These equations represent two different functions. If you rewrite the first equation in terms of y, you’ll find the equation of the function is y = 3x + 2. The correct answer is f(x) = 3x2 + 5 and y = 3x2 + 5. C) f(x) = 3x2 + 5 and y = 3x2 + 5Correct. The expressions that follow f(x) = and y = are the same, so these are two different ways to write the same function: f(x) = 3x2 + 5 and y = 3x2 + 5. D) None of the aboveIncorrect. Look at the expressions that follow f(x) = and y =. If the expressions are the same, then the equations represent the same exact function. The correct answer is f(x) = 3x2 + 5 and y = 3x2 + 5.
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iren2701 [21]

Answer:

445788

Step-by-step explanation:

OMG too difficult man!!

Thanks to my calculator to solve this hard question

5 0
2 years ago
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. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd
Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

x²y'' - 7xy' + 16y = 0; y1 = x^4

Answer:

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

x^6u'' + 8x^5u' + 12x^4u - 7x^5u' - 28x^4u + 16x^4u = 0

x^6u'' + x^5u' = 0

xu'' = -u'

Let w = u'

Then w' = u''

So

xw' = -w

w'/w = -1/x

Integrating both sides

lnw = -lnx + C

w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

6 0
3 years ago
What is y - 3y please ​
tatyana61 [14]

Answer:

<u>=-2y</u>

Step-by-step explanation:

y-3y

<u>=-2y</u>

5 0
3 years ago
Suppose in this semester, our Exam 1 average was about 86 with an SD of about 10. Suppose the correlation between our Exam 1 and
Oduvanchick [21]

Answer:

the slope of the regression equation for predicting our Exam 2 scores from Exam 1 scores is 0.492

And the y-intercept of the regression equation for predicting our Exam 2 scores from Exam 1 is 33.688

Step-by-step explanation:

Given the data in the question;  

mean X" = 86

SD σx = 10

Y" = 76

SD σy = 8.2

r = 0.6

Here, Exam 2 is dependent and Exam 1 is independent.

The Regression equation is

y - Y" = r × σy/σx  ( x - x" )

we substitute

y - 76 = 0.6 × 8.2/10  ( x - 86 )

y  - 76 = 0.492( x - 86 )

y  - 76 = 0.492x - 42.312

y = 0.492x - 42.312 + 76

y = 0.492x + 33.688

Hence, the slope of the regression equation for predicting our Exam 2 scores from Exam 1 scores is 0.492

And the y-intercept of the regression equation for predicting our Exam 2 scores from Exam 1 is 33.688

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3080 inches
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