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WARRIOR [948]
3 years ago
11

There are 250 students taking either band or chorus. If there are 180 students taking band and 60 students in both band and chor

us, how many students are only in chorus? 10

Mathematics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

<em>There are </em><em>70</em><em> students only in chorus.</em>

Step-by-step explanation:

Total number of students is 250

Out of 250, 180 students took band.

And 60 took band and chorus, i.e this is the intersection of the 2 groups.

The Venn diagram is attached below.

So, the number of students who took only band = 180-60 = 120

Total number of students or sum of the 3 regions is 250.

So, the number of students who took only chorus = 250-180 = 70

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Sladkaya [172]
The answer is the square root of 180
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A ball dropped from the top of a building can be modeled by the function f(t)=-16^2+36, where t represents time in seconds
Viefleur [7K]

Answer:

The ball is dropped from a height of 36 feet

The bee and the ball will collide after approximately 1.3 seconds

The bee and the ball will collide at approximately 8 feet above the ground

The ball hits the ground after 1.5 seconds

Step-by-step explanation:

<u><em>The complete question is</em></u>

A ball dropped from the top of the building can be modeled by the function f(t)=-16t^2 + 36 , where t represents time in seconds after the ball was dropped. A bee's flight can be modeled by the function, g(t)=3t+4, where t represents time in seconds after the bee starts the flight.

The graph represents the situation.

select all that apply

1) The bee launches into flight from the ground.

2) The ball is dropped from a height of 36 feet.

3) The bee and the ball will collide after approximately 1.3 seconds.

4) The bee and the ball will collide after approximately 8 seconds.

5) The bee and the ball will collide at approximately 8 feet above the ground.

6) The bee and the ball will not collide.

7) The ball hits the ground after 1.5 seconds

see the attached figure to better understand the problem

<u><em>Verify all the options</em></u>

case 1) The bee launches into flight from the ground.

The statement is false

Because

we have

g(t)=3t+4

For t=0

g(t)=3(0)+4=4\ ft

That means ----> The bee launches into flight from a height 4 feet above the ground

case 2) The ball is dropped from a height of 36 feet

The statement is true

Because

For t=0

f(t)=-16(0)^2+36=36\ ft

case 3) The bee and the ball will collide after approximately 1.3 seconds

The statement is true

Because

Equate f(t) and g(t)

-16t^2+36=3t+4

-16t^2-3t+32=0

solve the quadratic equation by graphing

The solution is t=1.324 sec

see the attached figure N 2

case 4) The bee and the ball will collide after approximately 8 seconds

The statement is false

Because, the bee and the ball will collide after approximately 1.3 seconds (see case 3)

case 5) The bee and the ball will collide at approximately 8 feet above the ground

The statement is true

Because

For t=1.324 sec

substitute the value of t in f(t) or g(t)

g(t)=3(1.324)+4=7.97\ ft

case 6) The bee and the ball will not collide

The statement is false

see case 3) and case 5)

case 7) The ball hits the ground after 1.5 seconds

The statement is true

Because

we know that

The ball hit the ground when the value of f(t) is equal to zero

so

For f(t)=0

-16t^2+36=0

t^2=\frac{36}{16}

t=\frac{6}{4}=1.5\ sec

4 0
3 years ago
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Given the following diagram, find the missing measure.<br> m&lt;1=30, m&lt;2=45, m&lt;3=
Mice21 [21]
I believer it is m3= 105
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3 years ago
Help me with these math questions....
Nady [450]

Answer: cotθ

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

 tanθ * cos²θ * csc²θ

=  \dfrac{sin\theta}{cos\theta} * \dfrac{cos\theta*cos\theta}{} *\dfrac{1}{sin\theta*sin\theta}

= \dfrac{cos\theta}{sin\theta}

= cotθ

Answer: B

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

The parent graph is y = x²

The new graph y = -x² + 3 should have the following:

  • reflection over the x-axis
  • vertical shift up 3 units

Answers:

  • a. Quadrant II
  • b. negative
  • c. \dfrac{\pi}{6}
  • d. C
  • e.-\dfrac{\sqrt{3}}{3}

<u>Explanation:</u>

\dfrac{17\pi}{6} - \dfrac{12\pi}{6} = \dfrac{5\pi}{6}

a) Quadrant 2 is: \dfrac{\pi}{2} < \theta < \pi

b) In Quadrant 2, cos is negative and sin is positive, so tan is negative

c) \pi-\dfrac{5\pi}{6} = \dfrac{\pi}{6}

d) the reference line is above the x-axis so it is negative -->  -tan\dfrac{\pi}{6}

e) tan(\dfrac{5\pi}{6})=\dfrac{1}{-\sqrt{3}}=-\dfrac{\sqrt{3}}{3}


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3 years ago
Question 1-6
Mumz [18]

i mean ig but 129.... ehh

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3 years ago
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