All categories

3 years ago

Explain what it means for a solution of an equation to be extraneous.

Mathematics

1 answer:

HACTEHA [7]3 years ago

4 0

Answer:

Extraneous Solutions An extraneous solution is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation. Example 1: Solve for x, 1 x − 2 + 1 x + 2 = 4 (x − 2) (x + 2).

You might be interested in

Which list shows the numbers in order from least to greatest?

Helga [31]

The one selected shows from greatest to least.

7 0

3 years ago

Read 2 more answers

I’m stuck, HELP<br><br> 3(6) + 4(2)^3 - 5

Slav-nsk [51]

Answer:

Step-by-step explanation:

Lets split the problem then add/subtract at the end

3 x 6 = 18

Due to Order of operations we do 2^3 first then multipy by 4

so 2^3 = 8 Then we do 8 x 4 = 32

Now we add/subtract

18 + 32 - 5 = 45

45 is the answer

4 0

3 years ago

Read 2 more answers

Salvador Dali was born on May 11, 1904. He died on January 23, 1989. Antoni Gaudi was born

Reika [66]

Answer:

22 yrs old (22 yrs & 29 days Old to be exact, because he died in June)

Step-by-step explanation:

Subtract:

1926 - 1904 = 22

Since Gaudi died on June 10th 1926 and Dali was born on May 11th, 1904, you just need to subtract the years.

8 0

3 years ago

The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris a

nadezda [96]

Answer and explanation:

Given : The probabilities of poor print quality given no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0, 0.3, 0.4, and 0.6, respectively.

The probabilities of no printer problem, misaligned paper, high ink viscosity, or printer-head debris are 0.8, 0.02, 0.08, and 0.1, respectively.

Let the event E denote the poor print quality.

Let the event A be the no printer problem i.e. P(A)=0.8

Let the event B be the misaligned paper i.e. P(B)=0.02

Let the event C be the high ink viscosity i.e. P(C)=0.08

Let the event D be the printer-head debris i.e. P(D)=0.1

and the probabilities of poor print quality given printers are

$P(E|A)=0,\ P(E|B)=0.3,\ P(E|C)=0.4,\ P(E|D)=0.6$

First we calculate the probability that print quality is poor,

$P(E)=P(A)P(E|A)+P(B)P(E|B)+P(C)P(E|C)+P(D)P(E|D)$

$P(E)=(0)(0.8)+(0.3)(0.02)+(0.4)(0.08)+(0.6)(0.1)$

$P(E)=0+0.006+0.032+0.06$

$P(E)=0.098$

a. Determine the probability of high ink viscosity given poor print quality.

$P(C|E)=\frac{P(E|C)P(C)}{P(E)}$

$P(C|E)=\frac{0.4\times 0.08}{0.098}$

$P(C|E)=\frac{0.032}{0.098}$

$P(C|E)=0.3265$

b. Given poor print quality, what problem is most likely?

Probability of no printer problem given poor quality is

$P(A|E)=\frac{P(E|A)P(A)}{P(E)}$

$P(A|E)=\frac{0\times 0.8}{0.098}$

$P(A|E)=\frac{0}{0.098}$

$P(A|E)=0$

Probability of misaligned paper given poor quality is

$P(B|E)=\frac{P(E|B)P(B)}{P(E)}$

$P(B|E)=\frac{0.3\times 0.02}{0.098}$

$P(B|E)=\frac{0.006}{0.098}$

$P(B|E)=0.0612$

Probability of printer-head debris given poor quality is

$P(D|E)=\frac{P(E|D)P(D)}{P(E)}$

$P(D|E)=\frac{0.6\times 0.1}{0.098}$

$P(D|E)=\frac{0.06}{0.098}$

$P(D|E)=0.6122$

From the above conditional probabilities,

The printer-head debris problem is most likely given that print quality is poor.

3 0

3 years ago

Read 2 more answers

Need help with this one

Anon25 [30]

ANSWER

$y = \frac{10}{3} x - 18$

EXPLANATION

The given line has slope

$m_{1} = - 0.3$

The line that is perpendicular to this line has a slope that is the negative reciprocal of -0.3

$m_2 = \frac{ - 1}{ - 0.3}$

$m_2 = \frac{10}{ 3}$

If the line e passes through (3,-8), then the equation is given by,

$y - - 8 = \frac{10}{3} (x - 3)$

$y = \frac{10}{3} x - 10 - 8$

$y = \frac{10}{3} x - 18$

3 0

3 years ago