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3 years ago

How many solutions are there to this equation 3x(x-4)+5-x=2x-7

Mathematics

1 answer:

stich3 [128]3 years ago

4 0

Answer:

2 solutions

Step-by-step explanation:

I like to use a graphing calculator to find solutions for equations like these. The two solutions are ...

x = 1
x = 4

To solve this algebraically, it is convenient to subtract 2x-7 from both sides of the equation:

3x(x -4) +5 -x -(2x -7) = 0

3x^2 -12x +5 -x -2x +7 = 0 . . . . . eliminate parentheses

3x^2 -15x +12 = 0 . . . . . . . . . . . . collect terms

3(x -1)(x -4) = 0 . . . . . . . . . . . . . . . factor

The values of x that make these factors zero are x=1 and x=4. These are the solutions to the equation. There are two solutions.

<em>Alternate method</em>

Once you get to the quadratic form, you can find the number of solutions without actually finding the solutions. The discriminant is ...

d = b^2 -4ac . . . . where a, b, c are the coefficients in the form ax^2+bx+c

d = (-15)^2 -4(3)(12) = 225 -144 = 81

This positive value means the equation has 2 real solutions.

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Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

3 years ago

The following data is from a simple random sample of 12 students GPAs. 3.12 2.45 4.0 3.76 3.54 2.78 3.39 3.21 1.98 3.43 3.98 2.7

Rufina [12.5K]

Answer: 3. 2

Step-by-step explanation:

Given the data:

3.12 2.45 4.0 3.76 3.54 2.78 3.39 3.21 1.98 3.43 3.98 2.77

Point estimate of population mean :

Σx / N

(3.12 + 2.45 + 4.0 + 3.76+ 3.54 + 2.78 + 3.39 + 3.21 + 1.98 + 3.43 + 3.98 + 2.77)

= 38.41 / 12

= 3.2008

4 0

3 years ago

A two-factor between-subjects design is evaluated. The F-value for Factor A has df = 1, 40 and the F-value for Factor B has df =

finlep [7]

Answer:

The df values for the A x B interaction are also 3,40

Step-by-step explanation:

The F-value for Factor A has df = 1, 40

The F-value for Factor B has df = 3, 40.

The df values for the A x B interaction are also 3,40

This is because the source of variation between columns has ( c- 1); rc(n-1) degrees of freedom and the source of variation between rows has (r-1); rc(n-1) degrees of freedom and the source of variation of interaction

(between rows and columns) has ( c-1) ( r-1); rc(n-1) degrees of freedom.

Therefore

Factor A: ( c- 1); rc(n-1) = 1,40

<u>Factor B: (r-1); rc(n-1) = 3,40</u>

Interaction : ( c-1) ( r-1); rc(n-1) = 3*1,(40)= 3,40

where c refers to columns , r refers to rows and n refers to the total sample size.

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3 years ago

Which inequality would result in the shaded solution on the unit circle to the right?

omeli [17]

<h3>Answer: Choice B</h3>

Explanation:

Cosine is positive in quadrants I and IV, but quadrant IV isn't shaded in so we can rule out choice A.

Sine is positive in quadrants I and II. So far it looks like choice B could work. In fact, it's the answer because sin(pi/6) = 1/2 and sin(5pi/6) = 1/2. So if 0 ≤ sin(x) < 1/2, then we'd shade the region between theta = 0 and theta = pi/6; as well as the region from theta = 5pi/6 to theta = pi.

Choice C is ruled out because tangent is positive in quadrants I and III, but quadrant III isn't shaded.

Choice D is ruled out for similar reasoning as choice A. Recall that $\sec(x) = \frac{1}{\cos(x)}$

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Serggg [28]

Answer: 7 Courts

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3 years ago

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