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sattari [20]
3 years ago
8

Find an equation of the hyperbola having foci at (-9,-4) and (7,-4) and vertices at (-6, -4) and (4.-4).​

Mathematics
1 answer:
Veronika [31]3 years ago
6 0

Answer:

-9,-4

Step-by-step explanation:

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On a single roll of a pair of dice, what are the odds against rolling a sum of 12?
o-na [289]

Answer:

\frac{1}{35}

Step-by-step explanation:

On a single roll of a pair of dice. When a pair of dice are rolled the possible outcomes are as follows:

(1,1)         (1,2)          (1,3)  (1,4)  (1,5)  (1,6)

(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)

(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)

(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)

(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)

(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6)

The number of outcomes that gives us 12 are (6,6). There is only one outcome that gives us sum 12.

Total outcomes = 36

Odd against favor = \frac{non \ favorable\ outcomes}{favorable \ outcomes}

Number of outcomes of getting sum 12 is 1

Number of outcomes of not getting sum 12 is 36-1= 35

odds against rolling a sum of 12= \frac{1}{35}

3 0
3 years ago
What rule would you use to translate a triangle 4 units to the right?
Novay_Z [31]

Answer:

here is what i found also you could of looked it up

Step-by-step explanation:

(x,y)→(x+4,y)

6 0
3 years ago
2x2 – 9x + 2 = –1
Reika [66]

Answer:

C)  The discriminant is greater than 0, so there are two real roots

Step-by-step explanation:

<u>Discriminant</u>

b^2-4ac\quad\textsf{when}\:ax^2+bx+c=0

\textsf{when }\:b^2-4ac > 0 \implies \textsf{two real roots}

\textsf{when }\:b^2-4ac=0 \implies \textsf{one real root}

\textsf{when }\:b^2-4ac < 0 \implies \textsf{no real roots}

2x^2-9x+2=-1

\implies 2x^2-9x+3=0

\implies a=2, b=-9, c=3

\begin{aligned}b^2-4ac &=(-9)^2-4(2)(3)\\& =81-24\\&=57 > 0\implies \textsf{two real roots}\end{aligned}

5 0
2 years ago
Grayson ran 2 1/2 miles around the track .each lap is 5/6 of a mile .how many laps did Grayson run?
andrezito [222]

Given:

Number of miles covered = 2\dfrac{1}{2}

Length of each lap = \dfrac{5}{6} of a mile.

To find:

The number of laps.

Solution:

We know that,

\text{Number of laps}=\dfrac{\text{Number of miles covered}}{\text{Length of each lap}}

\text{Number of laps}=\dfrac{2\dfrac{1}{2}}{\dfrac{5}{6}}

\text{Number of laps}=\dfrac{\dfrac{5}{2}}{\dfrac{5}{6}}

\text{Number of laps}=\dfrac{5}{2}\times \dfrac{6}{5}

\text{Number of laps}=3

Therefore, Grayson ran 3 laps.

5 0
2 years ago
Find the value of x in the triangle below.
geniusboy [140]

Answer:

Can't be Solved

Step-by-step explanation:

tan x = opp/adj

tan x = 23/11

tan x = 2.09090909 which is about

tan x = 2.09

x = 64.4 degrees which is the tangent of 2.09 in degrees.

x = 64.4 degrees

This is the only way to solve, since no answer is equivalent it can't be solved.

5 0
2 years ago
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