my teacher didn't explain well so I need help​

Answer:

1.01 m8

Step-by-step explanation:

Answer:

You sold 20 student tickets.

Step-by-step explanation:

Given that:

Total tickets sold = 27

Total amount collected = $170

Cost of student ticket = $5

Cost of adult ticket = $10

Let,

x be the number of students tickets sold

y be the number of adult tickets sold

x+y = 27          Eqn 1

5x+10y=170     Eqn 2

Multiplying Eqn 1 by 10

10(x+y=27)

10x+10y=270     Eqn 3

Subtracting Eqn 2 from Eqn 3

(10x+10y)-(5x+10y)=270-170

10x+10y-5x-10y=100

5x=100

Dividing both sides by 5

[tex]\frac{5x}{5}=\frac{100}{5}\\x=20[/tex]

Hence,

You sold 20 student tickets.

Answer:

V

≈

179.59in³

Step-by-step explanation:

Answer:

96000

Step-by-step explanation:

N=4years

R=4 %

We have S.I.=

100

PNR

=

100

P×4×4

=

100

16P

=0.16P

And on interest being compounded for 3 years and R=5 %, Amount=P(1+

100

R

)

N

=P(1+

100

5

)

3

=P×(1.05

3

)=1.157625P

So, C.I.=A−P=1.157625P−P=0.157625P

Given, S.I.−C.I=Rs228

=>0.16P−0.157625P=Rs228

=>0.002375P=Rs228

=>P=Rs96,000

Answer:

Splash Mountain = $7.25

Centrifugal Chamber= $5.25

Step-by-step explanation:

Click On Link For Work

Answer:

Either C or B I’m not quite sure which one. I’m sorry

Step-by-step explanation:

Answer:

3

Step-by-step explanation:

Took quiz

Answer:The answer is B cause the 0 repeats multiple times

Step-by-step explanation:

The slope of this line is 0. If you graph the line y=10, for any x value, the y-value remains 10 (denoted because x is equal to 0 in the actual equation shown, which means that x will not be shown in the equation). The graph would basically just be a straight horizontal line, neither curving up or down.

Answer: its B

Step-by-step explanation: i've done this test before

Answer:

the answer is b

Step-by-step explanation:

because i have done this question before

Answer:

The solutions to the system of equations are:

[tex]x=\frac{325}{9},\:y=\frac{65}{9}[/tex]

Step-by-step explanation:

Given the system of the equations

[tex]\begin{bmatrix}2x-y=65\\ 5y=x\end{bmatrix}[/tex]

isolate 'x' for 2x-y

[tex]2x-y=65[/tex]

Add y to both sides

[tex]2x-y+y=65+y[/tex]

[tex]2x=65+y[/tex]

Divide both sides by 2

[tex]\frac{2x}{2}=\frac{65}{2}+\frac{y}{2}[/tex]

[tex]x=\frac{65+y}{2}[/tex]

[tex]\mathrm{Subsititute\:}x=\frac{65+y}{2}[/tex]

isolate 'y' for  [tex]5y=\frac{65+y}{2}[/tex]

[tex]5y=\frac{65+y}{2}[/tex]

[tex]10y=65+y[/tex]

subtract y from both sides

[tex]10y-y=65+y-y[/tex]

[tex]9y=65[/tex]

Divide both sides by 9

[tex]\frac{9y}{9}=\frac{65}{9}[/tex]

[tex]y=\frac{65}{9}[/tex]

[tex]\mathrm{For\:}x=\frac{65+y}{2}[/tex]

[tex]\mathrm{Subsititute\:}y=\frac{65}{9}[/tex]

[tex]x=\frac{65+\frac{65}{9}}{2}[/tex]

[tex]x=\frac{325}{9}[/tex]

The solutions to the system of equations are:

[tex]x=\frac{325}{9},\:y=\frac{65}{9}[/tex]

Answer:

to be entirely honest i do not know

Step-by-step explanation:

Answer:

2x + 3y - 9 = 0

Step-by-step explanation:

17/20 = t + (- 13/20)

t = 17/20 + 13/20

t = 30/20

t = 1.5

Answer:

X=7

Step-by-step explanation:

Hope this helps, have a good day

Answer:

Equation 2

Step-by-step explanation:

Equation 2 is the only one possible to be linear. All of the other answers have an exponents. For instance, say you have y=x^3. For 1, it is 1*1*1. (1,1). For 2, it is 2*2*2. (2,8). For 3, 3*3*3. (3,27) It escalates into a curve and is non-linear.

Answer:

equation form: 12 + 2x = 18

x = 3

lets check the work see if its true

12 + 2(3) = 18

12 + 6 = 18

It Checks!

therefore your answer is 3. He can get 3 different toppings

Step-by-step explanation:

Answer:

x = infinite amount of solutions

General Formulas and Concepts:

Pre-Algebra

Step-by-step explanation:

Step 1: Define equation

3(x + 10) + 6 = 3(x + 12)

Step 2: Solve for x

Here we see that x does indeed equal x. We can plug in any number x and it would render the equation true.

∴ x = infinite amount of solutions

Answer:

infinitely many solutions / all solutions

Step-by-step explanation:

Given:

In a geometric sequence,

First term = 11

Common ratio = 4

To find:

The function for the geometric sequence.

Solution:

We have,

First term: a = 11

Common ratio: r = 4

We know that, nth term of a geometric sequence is

[tex]a_n=ar^{n-1}[/tex]

Putting a=11 and r=4, we get

[tex]a_n=(11)(4)^{n-1}[/tex]

The function notation is

[tex]f(n)=(11)(4)^{n-1}[/tex]

Therefore, the function for the geometric sequence is [tex]f(n)=(11)(4)^{n-1}[/tex], where n is a positive integer.

Answer:

i think its B but i'm not 100 percent sure sorry if im not but i think i am

Step-by-step explanation:

Answer:

It's D

Step-by-step explanation:

You l*w is the area.

Answer: H

Step-by-step explanation:

Answer:

0.7

Step-by-step explanation:

To get 7/10 as a decimal you simply divide 7 by 10 which gets you 0.7.

Answer:

h = — 10.071

Step-by-step explanation:

Look at the picture for my work.

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

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

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

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

This value is approximate. Round however you need to.

Step-by-step explanation:

3 ( _3 ) 2 + 5 ( -3 ) - 14 = _2

.........