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Evgen [1.6K]
3 years ago
8

You roll a fair six sided die what is P

Mathematics
1 answer:
-Dominant- [34]3 years ago
7 0

Answer:

<em>The answer is down below </em>

<em>~hope this helped~</em>

You might be interested in
|-9| = |9| true or false.
bogdanovich [222]

Step-by-step explanation:

1) -9 = 9 true/false

the answer is false because negative and positive quite unlike in terms.

2) -(-21) = -21

Using the symbols multiplication rule,

Where,

<h2>+ × + = +</h2><h2>- × - = +</h2><h2>- × + = -</h2><h2>+ × - = -</h2>

Therefore, -(-21) = +21 = 21

⟼ 21 = -21 \sf true / \:  \sout{false}

6 0
2 years ago
Read 2 more answers
Please help I have an exam tomorrow!!!!
Zepler [3.9K]
I think the most appropriate answer would be "-6.58".

I hope it helped you!
5 0
3 years ago
What is (2/3) to the 4th power?
Wewaii [24]
(\frac{2}{3})^4= \frac{2^4}{3^4}= \boxed{\frac{16}{81}}
8 0
3 years ago
Click on all the points that are solutions to y = x^2 - 9.
Oliga [24]

9514 1404 393

Answer:

  B) (2, -5)

  D) (3, 0)

Step-by-step explanation:

I find it convenient to let a graphing calculator plot the points and the graph.

The only two points on the graph of the curve are ...

  (2, -5) and (3, 0)

3 0
3 years ago
Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).
fomenos

Answer:

\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is 2a=|4--10|.

\implies 2a=|14|

\implies 2a=14

\implies a=7

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is (\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)

We need to use the relation a^2+b^2=c^2 to find b^2.

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

c=|6--3|=9

\implies 7^2+b^2=9^2

\implies b^2=9^2-7^2

\implies b^2=81-49

\implies b^2=32

We substitute these values into the standard equation of the hyperbola to obtain:

\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1

\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1

7 0
3 years ago
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