How many total possible outcomes are there on a standard 6 sided die?

Y = -x^2 + 4x<br><br> quadratic formula

marin [14]

Answer:

Graph the parabola using the direction, vertex, focus, and axis of symmetry.

Direction: Opens Down

Vertex:

( 2 , 4 )

Focus:

( 2 , 15 /4 )

Axis of Symmetry:

x = 2

Directrix:

y = 17 /4

x y

0 0

1 3

2 4

3 3

4 0

To find the x-intercept, substitute in

0 for y and solve for x . To find the y-intercept, substitute in 0 for x and solve for y .

x-intercept(s): ( 0 , 0 ) , ( − 4 , 0 )

y-intercept(s): ( 0 , 0 )

Step-by-step explanation:

4 0

3 years ago

If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

zmey [24]

Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

4 0

3 years ago

Read 2 more answers

Which expression is equivalent to \frac{10q^5w^7}{2w^3}.\ \frac{4\left(q6\right)^2}{w^{-5}} for all values of q and w where the

mestny [16]

Answer:

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9$

Step-by-step explanation:

Given

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

Required

Determine the equivalent expression

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

Simplify the first fraction

$\frac{5q^5w^7}{w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

Apply law of indices on the first fraction;

$5q^5w^{7-3}.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

$5q^5w^4.\ \frac{4\left(q^6\right)^2}{w^{-5}}$

$\ \frac{5q^5w^4*4\left(q^6\right)^2}{w^{-5}}$

Apply law of indices:

$5q^5w^4*4\left(q^6\right)^2 * w^{5}$

Evaluate the bracket

$5q^5w^4*4 * q^{12} * w^{5}$

Collect Like Terms

$5*4q^5* q^{12}*w^4 * w^{5}$

$20q^5* q^{12}*w^4 * w^{5}$

$20q^{5+12}*w^{4+5}$

$20q^{17}w^9$

Hence:

$\frac{10q^5w^7}{2w^3}.\ \frac{4\left(q^6\right)^2}{w^{-5}} =20q^{17}w^9$

7 0

3 years ago

What is the slope of the line that passes through the points (6,−10) and (3, -13) ?

vodka [1.7K]

Answer:

m = 1

Step-by-step explanation:

m = (-13 - - 10) / (3 - 6)

m = -3/-3

m = 1

8 0

2 years ago

There are 47 numbers Find the probability that i will pick the 6 winning numbers in a draw

Agata [3.3K]

Answer: You have six numbers on your ticket. In drawing the first number from the collection of 45, one of those six numbers must be drawn in order for you to still have a chance of winning the big prize. So following the draw, the probability is only 6/45 (or 1/9) that you’re still in the game.

You can now see that on each successive draw the probability steadily drops- to 5/44, 4/43, 3/42, 2/41, and for the last number 1/40. As other answers have stated, there are 8,145,060 possible draws of six numbers and only one makes you a winner.

If you bought ten tickets for every drawing, the odds reach a 50% chance of having won at about 520,550 drawings. Assuming two drawings per week, the chance of winning reaches 50% in about 5,200 years.

Of course, SOMEONE will win and if you don’t play, your chance of winning is zero. It’s kind of fun to occasionally buy lottery tickets just for the dream of winning but don’t spend a lot of money on them.

5.9K viewsView upvotes · Answer requested by Arthur C. Thorpe

7

How many total possible outcomes are there on a standard 6 sided die?​

How many total possible outcomes are there on a standard 6 sided die?