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Leokris [45]
4 years ago
14

Jose is applying to college. He receives information on 7 different colleges. He will apply to all of those he likes. He may lik

e none of them, all of them, or any combination of them. How many possibilities are there for the set of colleges that he applies to?
Mathematics
1 answer:
marta [7]4 years ago
6 0

Answer:

128 posibilities

Step-by-step explanation:

We have 7 colleges (A,B,C,...,H) which form a set with seven elements.

What you are asking is the number of elements (or cardinality) of the set that contains all possible sets formed by those 7 elements (or the "power set").

It is known that if n is the number of elements of a given set X, then the cardinality of the power set is 2^n.

Therefore, there are 2^7 or 128 possibilities (or elements) for the set of colleges that he applies to.

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Part 1: Identify the vertex, focus, and directrix of each. Then sketch the graph.
Lina20 [59]

Answer:  \bold{1.\quad \text{Vertex}:(4,-1)\qquad \text{Focus}:\bigg(4,-\dfrac{9}{8}\bigg)\qquad \text{Directrix}:y=-\dfrac{7}{8}}

                \bold{2.\quad \text{Vertex}:(2,1)\qquad \text{Focus}:\bigg(\dfrac{9}{4},1\bigg)\qquad \text{Directrix}:y=\dfrac{7}{4}}            

<u>Step-by-step explanation:</u>

The vertex form of a parabola is y = a(x - h)² + k  or  x = a(y - k)² + h

  • (h, k) is the vertex
  • p is the distance from the vertex to the focus
  • -p is the distance from the vertex to the directrix

    \bullet\quad a=\dfrac{1}{4p}

1) y = -2(x - 4)² - 1     →      a = -2   (h, k) = (4, -1)

a=\dfrac{1}{4p}\quad \rightarrow \quad -2=\dfrac{1}{4p}\quad \rightarrow \quad p=-\dfrac{1}{8}\\\\\text{Focus = Vertex + p}\\\\.\qquad = \dfrac{-8}{8}+\dfrac{-1}{8}\\\\.\qquad =-\dfrac{9}{8}\qquad \rightarrow \qquad \text{Focus}=\bigg(4,-\dfrac{9}{8}\bigg)\\\\\\\text{Directrix: y=Vertex - p}\\\\.\qquad \qquad y=\dfrac{-8}{8}-\dfrac{-1}{8}\\\\.\qquad \qquad y=-\dfrac{7}{8}

*******************************************************************************************

2) x = (y - 1)² + 2     →      a = 1   (h, k) = (2, 1)

a=\dfrac{1}{4p}\quad \rightarrow \quad 1=\dfrac{1}{4p}\quad \rightarrow \quad p=-\dfrac{1}{4}\\\\\text{Focus = Vertex + p}\\\\.\qquad = \dfrac{8}{4}+\dfrac{1}{4}\\\\.\qquad =\dfrac{9}{4}\qquad \rightarrow \qquad \text{Focus}=\bigg(\dfrac{9}{4},1\bigg)\\\\\\\text{Directrix: x=Vertex - p}\\\\.\qquad \qquad x=\dfrac{8}{4}-\dfrac{1}{4}\\\\.\qquad \qquad x=\dfrac{7}{4}

6 0
3 years ago
Can someone help me with this question
d1i1m1o1n [39]

Answer:

Angle 1 and angle 4 are congruent because all vertical angles are congruent. Angle 1 and angle 5 are congruent because corresponding angles are congruent. Idk the rest.

Step-by-step explanation:

7 0
3 years ago
How many ways can a person toss a coin 13 times so that the number of tails is between 7 and 11 inclusive?
lilavasa [31]
What are the choices?
4 0
4 years ago
Tom went with his dad to a shooting range. His dad bought him 10 bullets. For every time he got bullseye, the father gave his so
marusya05 [52]

Answer:  Tom hit the bullseye 23 times .

Step-by-step explanation:

Let the number of times Tom hit the bullseye be 'x'

Since we have given that

Total number of times Tom shot = 55

Number of times he missed the shot be '55-x'

Number of bullets he already have = 10

According to question, For every time, he got bullseye, the father gave him an extra bullet and after every miss he took one away.

At last he ran out of bullets.

So, it becomes,

10+1\times x-1(55-x)=0\\\\10+x-55+x=0\\\\2x-45=0\\\\2x=45\\\\x=\frac{45}{2}=22.5\\\\\text{ But number of times can't be in decimal }\\\\So,\ x=23

Hence, Tom hit the bullseye 23 times .

7 0
3 years ago
Find the standard form of the equation of the hyperbola satisfying the given conditions: X intercept +/- 6; foci at (-10,0) and
uysha [10]

Answer:

\frac{x^{2}}{36} - \frac{y^{2}}{64}=1

Step-by-step explanation:

Given an hyperbola with the following conditions:

  • Foci at (-10,0) and (10,0)
  • x-intercept +/- 6;

The following holds:

  • The center is midway between the foci, so the center must be at (h, k) = (0, 0).
  • The foci are 10 units to either side of the center, so c = 10 and c^2 = 100
  • The center lies on the origin, so the two x-intercepts must then also be the hyperbola's vertices.

Since the intercepts are 6 units to either side of the center, then a = 6 and a^2 = 36.

Then, a^2+b^2=c^2\\b^2=100-36=64

Therefore, substituting a^2 = 36. and b^2=64 into the standard form

\frac{x^{2}}{a^2} - \frac{y^{2}}{b^2}=1\\We \: have:\\ \dfrac{x^{2}}{36} - \dfrac{y^{2}}{64}=1

4 0
3 years ago
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