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xxTIMURxx [149]
3 years ago
5

Six kids and two adults are going to the circus. Kid's tickets are on sale for only half the price of adult tickets. The total c

ost is $48. How much is one kids ticket? -how much is one adult ticket
Mathematics
1 answer:
avanturin [10]3 years ago
6 0

Answer:

One kids ticket is $4.80 and one adult ticket is $9.60

Step-by-step explanation:

Create a system of equations where k is the cost of a kids ticket and a is the cost of an adult ticket:

6k + 2a = 48

k = 1/2a

Solve by substitution by substituting the second equation into the first one:

6k + 2a = 48

6(1/2a) + 2a = 48

Simplify and solve for a:

3a + 2a = 48

5a = 48

a = 9.6

Find the cost of a kids ticket by dividing this by 2, since they are on sale for half the price of adult tickets.

9.6/2

= 4.8

One kids ticket is $4.80 and one adult ticket is $9.60

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China was isolated because of
konstantin123 [22]

Answer:

The geography of China isolated it from other cultures because there were the Himalayan Mountains, the Tibet-Qinghai Plateau, the Taklimakan Desert, and the Gobi Desert

7 0
2 years ago
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Which of the following graphs shows a parabola with a vertex of (1,-9) and solutions of (-2,0) and (4,0)?
Mariulka [41]

Answer:

Hello,

Step-by-step explanation:

Roots are -2 and 4

y=k*(x+2)(x-4)

Vertex = (1,-9) is a point of the parabola

-9=k*(1+2)(1-4) ==> k=1

Equation of the parabola is y=(x+2)(x-4)

But you don' t have given the graphs !!!!

3 0
2 years ago
The 13th term of a geometric sequence is 16, 384 and the first term is 4. What is the common ratio?
AleksAgata [21]

Answer:

\large \boxed{2}

Step-by-step explanation:

The formula for the nth term of a geometric sequence is

aₙ = a₁rⁿ⁻¹

In your geometric sequence, a₁ = 4 and a₁₃ = 16 384.

\begin{array}{rcl}16384 & = & 4r^{(13 - 1)}}\\16384 & = & 4r^{12}\\4096 & = & r^{12}\\3.6124 & = & 12 \log r\\0.30102 & = & \log r\\r & = & 10^{0.30102}\\ & = & \mathbf{2}\\\end{array}\\\text{The common ratio is $\large \boxed{\mathbf{2}}$}

Check:

\begin{array}{rcl}16384 & = & 4(2)^{12}\\16384 & = & 4(4096)\\16384 & = & 16384\\\end{array}

It checks.

7 0
2 years ago
10. The lengths of the electrical extension cords in a workshop are 6 ft, 8 ft, 25 ft, 8 ft, 12 ft, 50 ft, and 25 ft. What are t
Nadya [2.5K]
6 ft= 6x12in= 72 in.
8 ft= 8x12in= 96 in.
12ft=12x12in=144 in
25ft=25x12in=300 in
50ft=50x12in=600 in

72 in, 96 in, 96 in, 144 in, 300 in, 300 in, 600 in
Mean= (72+96+96+144+300+300+600)÷7
1608÷7 = 229.7 inches
Median=middle value in set = 144 inches
Mode= value(s) tha occur most often = 96 inches and 300 inches
Range=difference of largest and smallest values in set = 600-72= 528
Choice A
3 0
3 years ago
Read 2 more answers
Please prove this........​
Crazy boy [7]

Answer:  see proof below

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

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
3 years ago
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