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3 years ago

Solve the radical equation. m - 3 = √ 19 - 3m. Which is an extraneous solution to the radical equation?

Mathematics

2 answers:

Sphinxa [80]3 years ago

7 0

The answer is m=5 and m=-2. There is no extraneous solution for this equation.

kari74 [83]3 years ago

5 0

Answer:

The extraneous solution is m = -2

Step-by-step explanation:

This is normally solved by squaring both sides and solving the resulting quadratic.

$(3-m)^2=19-3m\\\\m^2-3m-10=0 \quad\text{subtract 19-3m}\\\\(m-5)(m+2)=0 \quad\text{factor}$

The values of m that make the factors zero are 5 and -2. For m=5, the equation becomes ...

5-3 = √(19-3·5) ⇒ 2 = √4 . . . . . true

For m = -2, the equation becomes ...

-2-3 = √(19-3(-2)) ⇒ -5 = √25 . . . . . not true

The extraneous solution is m = -2.

_____

Solving the equation graphically, it becomes readily apparent that the only solution is m=5. Extra work is required to see the extraneous solution is m=-2.

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Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more

Aleks [24]

Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1) (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2) (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3) (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c) + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

4 0

3 years ago

What is the surface area of the rectangular prism below?

Andrew [12]

Answer:

B ( 432 units^2)

Step-by-step explanation:

Total surface area= 2*(6*6)+ 4*(15*6)

There are 2 squares at each end of the same size so total area will be 2*(6*6)

If we look at the rectangles( bottom, top and the sides), we can see that all are of same dimensions and there are 4 of them so it's 4*(15*6)

In short, total surface area= 2*(6*6)+ 4*(15*6) = 432 units^2

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3 years ago

If the two dice are rolled 45 times, predict about how many times a sum of 9 will be rolled.

Liula [17]

Answer:

Step-by-step explanation:

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2 years ago

a web designer charges a $200 fee plus $50 per hour to build a website. which equation represents the total cost, y, to a custom

dezoksy [38]

Y = 50x + 200 <=== ur equation

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3 years ago

Read 2 more answers

Find the half range Fourier sine series of the function

worty [1.4K]

The full range is $-\pi$ (length $2L=2\pi$ ), so the half range is $L=\pi$ . The half range sine series would then be given by

$f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx$

where

$b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx$

Essentially, this is the same as finding the Fourier series for the function

$\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0$

Integrating by parts yields

$b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n$

So the half range sine series for this function is simply

$f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n$

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2 years ago