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2 years ago

Factor and solve the problems in the photo ……….. pleaseeeee help i haven’t done algebra in 2 years

Mathematics

1 answer:

vladimir1956 [14]2 years ago

5 0

Answer:

30. 2x²-x-1=0

or, 2x²-2x+x-1=0

or, 2x(x-1)+1(x-1)=0

or, (x-1)(2x+1)=0

so, x-1 = 0 or, x = 1

and 2x+1 = 0 or, 2x = -1 or, x = -1/2

so, x = 1 and 1/2

31. 3x²-14x=5

or, 3x²-14x-5=0

or, 3x²-15x+x-5=0

or, 3x(x-5)+1(x-5)=0

or, (x-5)(3x+1)=0

so, x-5=0 or, x=5

and 3x+1=0 or, 3x=-1 or, x=-1/3

so, x = 5 and -1/3

Answered by GAUTHMATH

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What I the slope of a line that is perpendicular to the line 2y-3x=8

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ANSWER

$- \frac{2}{3}$

EXPLANATION

The given given equation is

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We need to rewrite this equation in the slope-intercept form:

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We add 3x to both sides.

$2y - 3x + 3x=8 + 3x$

$\implies \: 2y = 3x + 8$

We divide through by 2 to get,

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The slope of this line is

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Let the slope of the line perpendicular to this line be 'n' .

Then the product of the slopes of two perpendicular lines is always negative 1.

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$n = - \frac{2}{3}$

Therefore the slope of the new line is

$- \frac{2}{3}$

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2 years ago

A point goes through the origin and has a slope of 3. What is the equation of the line?

Schach [20]

Answer:

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Step-by-step explanation:

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KiRa [710]

Answer:

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(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let X = number of text messages receive or send in an hour.

The random variable X follows a Poisson distribution with parameter λ.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: $\lambda=\frac{62.7}{24}= 2.6125$ .

The probability of a random variable can be computed using the formula:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...$

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

$P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180$

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667$

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

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