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1 year ago

4x^2 - y^2 - 2x- y Factorise each of the following expressions completely

Mathematics

1 answer:

Sergeu [11.5K]1 year ago

3 0

Answer:

(2x + y)(2x - y - 1)

Step-by-step explanation:

4x² - y² - 2x - y

4x² - y² is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , then

4x² - y²

= (2x)² - y²

= (2x - y)(2x + y)

expression is now

(2x - y)(2x + y) - (2x + y) ← factor out (2x + y) from each term

= (2x + y)(2x - y - 1)

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A building engineer analyzes a concrete column with a circular cross section. The circumference of the column is 18π meters.

Whitepunk [10]

Answer:

$A=81\pi\ m^{2}$

Step-by-step explanation:

Step 1

Find the radius of the circular cross section

we know that

The circumference is equal to

$C=2\pi r$

we have

$C=18\pi\ m$

substitute and solve for r

$18\pi=2\pi r$

simplify

$18=2 r$

$r=9\ m$

Step 2

Find the area of the circular cross section

The area of the circle is equal to

$A=\pi r^{2}$

we have

$r=9\ m$

substitute

$A=\pi (9)^{2}$

$A=81\pi\ m^{2}$

8 0

3 years ago

What is the slope-intercept form of the equation of the line with the slope of -5/8 that passes through the point (-14,6)

inessss [21]

Step-by-step explanation:

y-y1 = m(x-x1) is the equation for a linear line in y=mx+b (slope intercept)

slope = m = -5/8

x1 = -14

y1 = 6

y-6 = -5/8 (x--14)

y-6 = -5/8 (x+14)

y-6 = -5/8x-70/8

y-48/8= -5/8x-70/8

y = -5/8x - 22/8

When x = 0, y = -22/8, which makes (0, -22/8) your y intercept

5 0

3 years ago

Is 3/6 greater than 0/6

tekilochka [14]

Yes 3/6 is greater than 0/6.
0/6 is equal to 0

7 0

3 years ago

Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????

LenaWriter [7]

Answer:

$E[X^2]= \frac{2!}{2^1 1!}= 1$

$Var(X^2)= 3-(1)^2 =2$

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

$\phi(t) = E[e^{tX}]$

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

$\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx$

And we have that the moment generating function can be write like this:

$\phi(t) = e^{\frac{t^2}{2}$

And we can write this as an infinite series like this:

$\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...$

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

$E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...$

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

$\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}$

And then we have this:

$E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...$

And then we can find the $E[X^2]$

$E[X^2]= \frac{2!}{2^1 1!}= 1$

And we can find the variance like this :

$Var(X^2) = E[X^4]-[E(X^2)]^2$

And first we find:

$E[X^4]= \frac{4!}{2^2 2!}= 3$

And then the variance is given by:

$Var(X^2)= 3-(1)^2 =2$

7 0

3 years ago

The price of a stock rose from a yearly low of $9.68 to $18.61. what was the stock percent increase from its low price that year

Helga [31]

Answer: i think it is 8.93

Step-by-step explanation:

4 0

3 years ago