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3 years ago

Geometry help needed!

Mathematics

1 answer:

salantis [7]3 years ago

3 0

Answer:

AC Midpoint (2,0)-

CB Midpoint (5,-4)

midlength is 5

Step-by-step explanation:

midpoint formula $(\frac{x_{1}+x_{2} }{2} ,\frac{y_{1} +y_{2} }{2})$

A(2,4) C(2,-4)

$\frac{2+2}{2} =\frac{4}{2}=2\\\frac{4-4}{2} =\frac{0}{2} =0$

(2,0)

C(2,-4) B(8,-4)

$\frac{8+2}{2} =\frac{10}{2} =5$

$\frac{-4-4}{2} =\frac{-8}{2} =-4$ midpoint is (5,-4)

Distance between the 2 midpoints is the midsegment

(2,0) and (5,-4)

$d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1}) ^{2} }$

$d=\sqrt{(5-2)^{2} +(-4-0)^{2} } \\d=\sqrt{3^{2} +(-4^{2})\\ } \\d=\sqrt{x} 9+16\\d=\sqrt{25} \\d=5$

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2 years ago

Simplify; 1/x^2+5x+6 + 1/x^2+3x+2 pls I need a step by step solution urgently

ikadub [295]

Answer:

The simplest form of $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ is $\frac{2}{(x+1)(x+3)}$

Step-by-step explanation:

To add two algebraic fractions do these steps

Factorize each denominator
Simplify each fraction to its lowest term
Find the LCM of the two denominators
Divide LCM by each denominator and multiply the numerators by its corresponding quotients
Add the two fraction and simplify the last answer

To simplify $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$

Factorize each denominator

∵ The denominator of the first fraction is x² + 5x + 6

∵ x² = (x)(x)

∵ 6 = (2)(3)

∵ (2)(x) + (3)(x) = 5x ⇒ middle term

∴ x² + 5x + 6 = (x + 2)(x + 3)

∵ The denominator of the second fraction is x² + 3x + 2

∵ x² = (x)(x)

∵ 2 = (2)(1)

∵ (2)(x) + (1)(x) = 3x ⇒ middle term

∴ x² + 3x + 2 = (x + 2)(x + 1)

Find the LCM of the two denominators

∵ The denominators are (x + 2)(x + 3) and (x + 2)(x + 1)

- LCM is all the different factors multiplied together

∴ LCM of them is (x + 1)(x + 2)(x + 3)

- Divide LCM by each denominator

∵ (x + 1)(x + 2)(x + 3) ÷ (x + 2)(x + 3) = (x + 1)

- Multiply the numerator of the first fraction by (x + 1)

∵ (x + 1)(x + 2)(x + 3) ÷ (x + 2)(x + 1) = (x + 3)

- Multiply the numerator of the second fraction by (x + 3)

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{x+1}{(x+1)(x+2)(x+3)}+\frac{x+3}{(x+1)(x+2)(x+3)}$

- Add the numerators and write the answer as a single fraction

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{2x+4}{(x+1)(x+2)(x+3)}$

- Factorize the numerator by taking 2 as a common factor

∵ 2x + 4 = 2(x + 2)

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{2(x+2)}{(x+1)(x+2)(x+3)}$

- Simplify the fraction by dividing up and down by (x + 2)

∴ $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ = $\frac{2}{(x+1)(x+3)}$

The simplest form of $\frac{1}{x^{2}+5x+6}+\frac{1}{x^{2}+3x+2}$ is $\frac{2}{(x+1)(x+3)}$

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3 years ago

What is the solution set of y = x2 + 2x + 7 and y = x + 7? A. {(0, 7), (-1, 6)} B. {(0, 7), (-7, 0)} C. {(0, 7), (1, 8)} D. {(-2

Elena-2011 [213]

Answer:

Step-by-step explanation:

$y=x^2+2x+7$ AND

y = x + 7, then by the transitive property of equality:

$x^2+2x+7=x+7$

We can solve for the values of x by getting everything on one side of the equals sign and then solving for x:

$x^2+x=0$

We can factor out the common x to get:

x(x + 1) = 0

which tells us by the Zero Product Property that either

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

We are expecting 2 solutions for x since this is a second degree polynomial. We will sub both -1 and 0 into y = x + 7 to solve for the corresponding values of y

y = 0 + 7 so

y = 7 and the coordinate is (0, 7)

y = -1 + 7 so

y = 6 and the coordinate is (-1, 6)

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3 years ago

Is the equation true,false, or open? 14+7+(-1)=21

yuradex [85]

Answer:

The Equation is False.

Step-by-step explanation:

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3 years ago

Read 2 more answers

The number of errors in a textbook follow a Poisson distribution with a mean of 0.03 errors per page. What is the probability th

maksim [4K]

Answer:

The probability that there are 3 or less errors in 100 pages is 0.648.

Step-by-step explanation:

In the information supplied in the question it is mentioned that the errors in a textbook follow a Poisson distribution.

For the given Poisson distribution the mean is p = 0.03 errors per page.

We have to find the probability that there are three or less errors in n = 100 pages.

Let us denote the number of errors in the book by the variable x.

Since there are on an average 0.03 errors per page we can say that

the expected value is, $\lambda$ = E(x)

= n × p

= 100 × 0.03

= 3

Therefore the we find the probability that there are 3 or less errors on the page as

P( X ≤ 3) = P(X = 0) + P(X = 1) + P(X=2) + P(X=3)

Using the formula for Poisson distribution for P(x = X ) = $\frac{e^{-\lambda}\lambda^X}{X!}$

Therefore P( X ≤ 3) = $\frac{e^{-3} 3^0}{0!} + \frac{e^{-3} 3^1}{1!} + \frac{e^{-3} 3^2}{2!} + \frac{e^{-3} 3^3}{3!}$

= 0.05 + 0.15 + 0.224 + 0.224

= 0.648

The probability that there are 3 or less errors in 100 pages is 0.648.

7 0

3 years ago

Read 2 more answers