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3 years ago

Which statement best describes the association between variable X and variable Y?

Mathematics

2 answers:

Serjik [45]3 years ago

8 0

Not all heroes wear capes. Anyways thank you!

Lerok [7]3 years ago

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Five star review. Would come again.

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the sum of kyras and calebs age is 37. calebs age is equal to twice the quantity of kyras age decreased by 4. let k represent ky

nasty-shy [4]

Caleb - c
Kyras - k

k+c=37
c=2k-4

k+(2k-4)=37
3k=41
k=41/3= ~13.67 years old

c=37-k
c=(111-41)/3
c=70/3=~23.33 years old

7 0

3 years ago

An equation of a hyperbola is given.

siniylev [52]

Answer:

The vertices are $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

The foci are $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$ .

The asymptotes are $y=2x,\:y=-2x$ .

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

$\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1$ is the standard equation for a right-left facing hyperbola with center $\left(h,\:k\right)$ .

The vertices $\:\left(h+a,\:k\right),\:\left(h-a,\:k\right)$ are the two bending points of the hyperbola with center $\:\left(h,\:k\right)$ and semi-axis a, b.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and vertices $\left(3,\:0\right),\:\left(-3,\:0\right)$ .

For a right-left facing hyperbola, the Foci (focus points) are defined as $\left(h+c,\:k\right),\:\left(h-c,\:k\right)$ where $c=\sqrt{a^2+b^2}$ is the distance from the center $\left(h,\:k\right)$ to a focus.

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ $c=\sqrt{3^2+6^2}= 3\sqrt{5}$ and foci $\left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)$

The asymptotes are the lines the hyperbola tends to at $\pm \infty$ . For right-left hyperbola the asymptotes are: $y=\pm \frac{b}{a}\left(x-h\right)+k$

Therefore,

$\frac{x^2}{9}-\frac{y^2}{36}=1$ , is a right-left Hyperbola with $\:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6$ and asymptotes

$y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x$

b) The length of the transverse axis is given by $2a$ . Therefore, the lenght is 6.

c) See below.

4 0

4 years ago

Solve the system of equations. 3x+4y=−23 x=3y+1

larisa86 [58]

Answer:

Step-by-step explanation:

3x+4y=−23.........(1)

x=3y+1 .........(2)

Putting (2) in (1)

3(3y + 1) + 4y = -23

9y + 3 + 4y = -23

9y + 4y = -23 - 3

13y = -26

y = -26/13

y = -2

Putting y into (2)

x=3y+1

x = 3(-2) + 1

x = -6 + 1

x = -5

7 0

3 years ago

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!

Valentin [98]

Answer:

Step-by-step explanation:

Slope-intercept equation of a line is,

y = mx + b

Here, m = Slope of the line

b = y-intercept

If m = -6 and b = 3,

Equation of the line will be,

y = -6x + 3

Input - output values got the graph of this line,

x -1 0 1 2

y 9 3 -3 -9

Now by plotting these points we can draw the line on graph.

7 0

3 years ago

Factor<br> x ^ 3 + 2x ^ 2 – 5x - 10 = 0

aleksklad [387]

Answer:

Step-by-step explanation:

When you have four terms and 3rd degree equation (the highest power is 3) you will want to try to "factor by grouping"

Group together two terms, watch out for the negative signs!

x^3 + 2x^2-5x-10=0

(x^3 + 2x^2) - (5x+10)=0

Find a common factor in each group and factor it out. You're hoping that what is left in the parenthesis is the same in both cases.

x^2(x+2)-5(x+2)=0

Now you can factor out that (x+2) because it is both terms.

x^2(x + 2) - 5(x + 2)=0

~~~~ ~~~~

Pull these out.

What will be left is the x^2 and the - 5 (dont lose that - in front of the 5)

(x + 2)(x^2 - 5) = 0

If all you have to do is factor, then you're done. It is factored. But if you have to "solve" also, then put x+2=0 and x^2-5=0 and solve.

x = -2 and x = +- sqrt5

7 0

2 years ago