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Ratling [72]
3 years ago
12

In one day a museum collected $1590 from 321 people. The price of admission is $6 for an adult and $4 for a child. How many adul

ts and how many children were admitted to the museum?
I NEED IN STEP BY STEP SO I CAN UNDERSTAND!!!!
Mathematics
1 answer:
Troyanec [42]3 years ago
6 0

Answer:

153 adults and 168 children

Step-by-step explanation:

x - number of adults

y - number of children

\left \{ {{6x + 4y = 1590} \atop {x + y = 321}} \right.

\left \{ {{6x+4y = 1590} \atop {x=321-y}} \right.

\left \{ {{6*(321 - y)+4y = 1590} \atop {x=321-y}} \right.

\left \{ {{1926 -6y+4y=1590} \atop {x=321-y}} \right.

\left \{ {{1926 -2y=1590} \atop {x=321-y}} \right.

\left \{ {{-2y=1590-1926} \atop {x=321-y}} \right.

\left \{ {{-2y=-336} \atop {x=321-y}} \right.

\left \{ {{y=168} \atop {x=321-y}} \right.

\left \{ {{y=168} \atop {x=321-168}} \right.

\left \{ {{y=168} \atop {x=153}} \right.

Is it all clear?

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Sum of two numbers is 24 and the difference is 15.What are the two numbers ?
Alexeev081 [22]
Let,
the bigger number  be "x"
the smaller number be "y"
Now, according to the question,
x + y = 24...........equation (1).............because the sum of numbers is 24
x - y = 15 .............equation (2)..................because their difference is 15

Taking equation (2),
x - y = 15
x = 15 + y ....................equation (3)

Now, Taking equation (1)
x + y = 24
(Substituting the value of "x" form equation (3) we get)
15 + y + y = 24
(Subtracting 15 on both sides, we get)
y + y = 24 - 15
2y = 9
y = 9 / 2
y = 4.5
Now, 
Taking equation (3)
x = 15 + y
(Substituting the value of y , we get)
x = 15 + 4.5
x = 19.5

So, the two numbers are 19.5 and 4.5

5 0
3 years ago
In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a
nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ};

\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ};

\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q.

Also,

\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P in right triangle PBM.

Thus \rm \angle{P\hat{B}M} = \angle C\hat{M}Q.

Additionally \rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}.

Therefore \rm \triangle PBM \sim \triangle MCQ.

\rm \displaystyle BC = 2\;MC for M is the midpoint of segment BC.

\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC.

\rm \triangle PBM \sim \triangle MCQ implies that

\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3.

8 0
3 years ago
F(x) =
slega [8]

Answer:

Step-by-step explanation:

the graph correspond to the function f :

6 0
3 years ago
What are the constants in the expression 7r - 3 + 2s + 5
slega [8]
B. -3 and 5

constants are the numbers without variables (the letters) and coefficients are the numbers with variables
6 0
3 years ago
Find the discriminant of x2-6x -10 = 0, and determine the number of real solutions of the equation.​
sergiy2304 [10]

Answer:

76

2 real solutions

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Algebra I</u>

Standard Form: ax² + bx + c = 0

Discriminant: b² - 4ac

  • Positive - 2 solutions
  • Equal to 0 - 1 solution
  • Negative - No solutions/Imaginary

Step-by-step explanation:

<u>Step 1: Define</u>

x² - 6x - 10 = 0

<u>Step 2: Identify Variables</u>

<em>Compare quadratic.</em>

x² - 6x - 10 = 0 ↔ ax² + bx + c = 0

a = 1, b = -6, c = -10

<u>Step 3: Find Discriminant</u>

  1. Substitute in variables [Discriminant]:                                                            (-6)² - 4(1)(-10)
  2. [Discriminant] Evaluate exponents:                                                                36 - 4(1)(-10)
  3. [Discriminant] Multiply:                                                                                    36 + 40
  4. [Discriminant] Add:                                                                                          76

This tells us that our quadratic has 2 real solutions.

8 0
3 years ago
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