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KonstantinChe [14]
2 years ago
14

I need help please. If you are correct, I will give you Brainliest.

Mathematics
1 answer:
earnstyle [38]2 years ago
5 0

Answer:

(C) 220

Step-by-step explanation:

Let x represent the number of adult tickets sold and y represent the number of student tickets sold. With the information given, we can set up two equations:

x+y=360

5x+3y=1360 (Since for every adult ticket sold, $5 is made and for every student ticket sold, $3 is made)

In the first equation, we can represent x in terms of y:

x=360-y

And then, we can substitute x in the second equation for 360 - y to get:

5(360-y)+3y=1360 which simplifies to:

1800-2y=1360 and therefore, y=220.

Hope this helps :)

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Nancy found that x = 1 is one solution to the quadratic equation (x + 2)2 = a. What is the value of a?
solniwko [45]

ANSWER

a=9

EXPLANATION

The given quadratic equation is

{(x + 2)}^{2}  = a

If x=1 is a solution, then it must satisfy this equation:

{(1 + 2)}^{2}  = a

{3}^{2}  = a

a = 9

Therefore the value of 'a' is 9

4 0
3 years ago
Read 2 more answers
A fair die is tossed 5 times. Let <img src="https://tex.z-dn.net/?f=%5Cdfrac%7Bm%7D%7Bn%7D" id="TexFormula1" title="\dfrac{m}{n}
Ray Of Light [21]

Answer:

m = 1

Step-by-step explanation:

We can suppose that the number we are looking for is for example 5.

(we can do so because the probability is the same for each number - it'sna fair dice)

For the first toss the probability we have 5 is 1/6 (we have 6 numbers on the dice and number 5 is just one of the possible 6 outcomes).

For the second toss the probability we have 5 is again 1/6.

For the rest of 3 tosses we don'tcare what number we will get( we have our two consecutive 5s), so all of the outcomes for the rest of 3 tosses are good for us (probability is 6/6 = 1)

Threfore, the probability to get two consecutive 5s is 1/6 * 1/6 * 1 * 1 * 1 = 1/36.

We can see that m = 1.

4 0
3 years ago
Can someone help me on this pls? It’s urgent, so ASAP (it’s geometry)
GarryVolchara [31]

<u>Question 6</u>

1) \overline{AB} \cong \overline{BD}, \overline{CD} \perp \overline{BD}, O is the midpoint of \overline{BD}, \overline{AB} \cong \overline{CD} (given)

2) \angle ABO, \angle ODC are right angles (perpendicular lines form right angles)

3) \triangle ABO, \triangle CDO are right triangles (a triangle with a right angle is a right triangle)

4) \overline{BO} \cong \overline{OD} (a midpoint splits a segment into two congruent parts)

5) \triangle ABO \cong \triangle CDO (LL)

<u>Question 7</u>

1) \angle ADC, \angle BDC are right angles), \overline{AD} \cong \overline{BD}

2) \overline{CD} \cong \overline{CD} (reflexive property)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \triangle ADC \cong \triangle BDC (LL)

5) \overline{AC} \cong \overline{BC} (CPCTC)

<u>Question 8</u>

1) \overline{CD} \perp \overline{AB}, point D bisects \overline{AB} (given)

2) \angle CDA, \angle CDB are right angles (perpendicular lines form right angles)

3) \triangle CDA, \triangle CDB are right triangles (a triangle with a right angle is a right triangle)

4) \overline{AD} \cong \overline{DB} (definition of a bisector)

5) \overline{CD} \cong \overline{CD} (reflexive property)

6)  \triangle ADC \cong \triangle BDC (LL)

7) \angle ACD \cong \angle BCD (CPCTC)

8 0
1 year ago
Janet set off on a bicycle trip at 7:30 a.m. at an average rate of 24 miles per hour. Her husband John promised to bring her som
Gemiola [76]

I believe the answer would be 10:20 am.

5 0
3 years ago
3. Find the volume of a
weeeeeb [17]

Answer: 401.88

Step-by-step explanation: math

4 0
3 years ago
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