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Alja [10]
3 years ago
13

a team sold 830 tickets, collecting 41,170 dollars.if regular seats costs 50 dollars and the cheap ones cost 20, how many cheap

tickets were sold?
Mathematics
1 answer:
goldfiish [28.3K]3 years ago
4 0
I want to say 11.5 i might be really close
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Which is the product of (1 +41+3x) (2 - 7x–9x)?​
Andrews [41]

Answer:

-48x^2 -666x+84

Step-by-step explanation:

Exponet is ^2.

Multiply is •

Use Distributed Property. Multiply the second parathesis by each term from the first parathesis.

Any expression multiplied by 1 remains the same. Distribute 41 through the parathesis. 41(2-7x-9x).

Distribute 3x through the parathesis 3x • (2-7x-9x).

You will have these numbers left: (Slide 2)

Add the numbers, 2 and 82 to get 84. Add -7x -9x + -287x -369x + 6x to get -666x. Add -21x^2 - 27x^2 to get 48x^2.

Use the communitative property to reorder the terms. And then your answer is -48x^2 -666x+84.

6 0
3 years ago
Please help me is about pi​
Over [174]

Answer:

circumference TO the diameter

Step-by-step explanation:

circumference of the circle TO the diameter of the circle

6 0
3 years ago
Read 2 more answers
The temperature of a chemical reaction ranges between -10 degrees Celcius and 50 degrees Celcius. The temperature is at its lowe
kherson [118]

Answer:

The cosine function is f(t) = -30sin(π÷3)t + 20

Step-by-step explanation:

Given : The temperature of a chemical reaction ranges between −10 degrees Celsius and 50 degrees Celsius. The temperature is at its lowest point when t = 0, and the reaction completes 1 cycle during a 6-hour period.

To find : What is a cosine function that models this reaction?

General form of cosine function is f(x) = A cos(Bx)+C

Where A is the amplitude

B=2π÷Period

C is the midline    

Now, We have given

The temperature of a chemical reaction ranges between −10 degrees Celsius and 50 degrees Celsius.

A is the average of temperature,

i.e,

A=(-10-50)÷2 = -30

Period of 1 cycle is 6 hour

So,

B = 2π÷6 = π÷3

The temperature is at its lowest point when t = 0 and we know lowest point is -10

So,

f(t) = A cos t + C

-10 = -30 cos 0 + C

Therefore, C = 20

Substituting the values, we get

The cosine function is f(t) = -30sin(π÷3)t + 20

Read more at:

brainly.com/question/4411386

#SPJ10

8 0
2 years ago
Which equation is y = 2x2 – 8x 9 rewritten in vertex form? y = 2(x – 2)2 9 y = 2(x – 2)2 5 y = 2(x – 2)2 1 y = 2(x – 2)2 17
saveliy_v [14]

Answer:

Step-by-step explanation:

we have

y=2x2-8x+9

Group terms that contain the same variable, and move the constant to the opposite side of the equation

y-9=2x2-8x

y-9=2(x2-4x)

y-9+8=2(x2-4x+4)

y-1=2(x2-4x+4)

y-1=2(x-2)2

y=2(x-2)2+1

the answer is the option C

y=2(x-2)2+1

3 0
2 years ago
Will mark brainliest for the correct answer!
romanna [79]

Part (a)

Focus on triangle PSQ. We have

angle P = 52

side PQ = 6.8

side SQ = 5.4

Use of the law of sines to determine angle S

sin(S)/PQ = sin(P)/SQ

sin(S)/(6.8) = sin(52)/(5.4)

sin(S) = 6.8*sin(52)/(5.4)

sin(S) = 0.99230983787513

S = arcsin(0.99230983787513)

S = 82.889762826274

Which is approximate

------------

Use this to find angle Q. Again we're only focusing on triangle PSQ.

P+S+Q = 180

Q = 180-P-S

Q = 180-52-82.889762826274

Q = 45.110237173726

Which is also approximate.

A more specific name for this angle is angle PQS, which will be useful later in part (b).

------------

Now find the area of triangle PSQ

area of triangle = 0.5*(side1)*(side2)*sin(included angle)

area of triangle PSQ = 0.5*(PQ)*(SQ)*sin(angle Q)

area of triangle PSQ = 0.5*(6.8)*(5.4)*sin(45.110237173726)

area of triangle PSQ = 13.0074347717966

------------

Next we'll use the fact that RS:SP is 2:1.

This means RS is twice as long as SP. Consequently, this means the area of triangle RSQ is twice that of the area of triangle PSQ. It might help to rotate the diagram so that line PSR is horizontal and Q is above this horizontal line.

We found

area of triangle PSQ = 13.0074347717966

So,

area of triangle RSQ = 2*(area of triangle PSQ)

area of triangle RSQ = 2*13.0074347717966

area of triangle RSQ = 26.0148695435932

------------

We're onto the last step. Add up the smaller triangular areas we found

area of triangle PQR = (area of triangle PSQ)+(area of triangle RSQ)

area of triangle PQR = (13.0074347717966)+(26.0148695435932)

area of triangle PQR = 39.0223043153899

------------

<h3>Answer: 39.0223043153899</h3>

This value is approximate. Round however you need to.

===========================================

Part (b)

Focus on triangle PSQ. Let's find the length of PS.

We'll use the value of angle Q to determine this length.

We'll use the law of sines

sin(Q)/(PS) = sin(P)/(SQ)

sin(45.110237173726)/(PS) = sin(52)/(5.4)

5.4*sin(45.110237173726) = PS*sin(52)

PS = 5.4*sin(45.110237173726)/sin(52)

PS = 4.8549034284642

Because RS is twice as long as PS, we know that

RS = 2*PS = 2*4.8549034284642 = 9.7098068569284

So,

PR = RS+PS

PR = 9.7098068569284 + 4.8549034284642

PR = 14.5647102853927

-------------

Next we use the law of cosines to find RQ

Focus on triangle PQR

c^2 = a^2 + b^2 - 2ab*cos(C)

(RQ)^2 = (PR)^2 + (PQ)^2 - 2(PR)*(PQ)*cos(P)

(RQ)^2 = (14.5647102853927)^2 + (6.8)^2 - 2(14.5647102853927)*(6.8)*cos(52)

(RQ)^2 = 136.420523798282

RQ = sqrt(136.420523798282)

RQ = 11.6799196828694

--------------

We'll use the law of sines to find angle R of triangle PQR

sin(R)/PQ = sin(P)/RQ

sin(R)/6.8 = sin(52)/11.6799196828694

sin(R) = 6.8*sin(52)/11.6799196828694

sin(R) = 0.4587765387107

R = arcsin(0.4587765387107)

R = 27.3081879220073

--------------

This leads to

P+Q+R = 180

Q = 180-P-R

Q = 180-52-27.3081879220073

Q = 100.691812077992

This is the measure of angle PQR

subtract off angle PQS found back in part (a)

angle SQR = (anglePQR) - (anglePQS)

angle SQR = (100.691812077992) - (45.110237173726)

angle SQR = 55.581574904266

--------------

<h3>Answer: 55.581574904266</h3>

This value is approximate. Round however you need to.

8 0
3 years ago
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