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NeX [460]
3 years ago
8

The student council is planning for the prom. They have broken down their expenses in the graph below. If they decide they will

spend $1,200 on refreshments, then what is their total budget?
Prom Budget:
Entertainment: 25%
Refreshments: 30%
Decorations: 45%
Answer choices:
A. $3,600
B. $6,000
C. $7,200
D. $4,000
(Please help me!)
Mathematics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

D

Step-by-step explanation:

1200=30%

1200/x=30%

1200*.3=400

400=x

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The answer is D because there is two completely different events going on, and they have nothing to do with each other.
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4 years ago
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Which perfect square equation represents x^2+4x+1=0 by completing the square?
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7 0
3 years ago
A gym initially has 5000 members. Each year, the gym loses 10% of its current members and gains 100 new members. Which recursive
Elis [28]

Answer:

An = A0 + 100 * n - (An-1) * 0.10

Step-by-step explanation:

Tenemos una regla recursiva para una secuencia es una fórmula que nos dice cómo avanzar de un término a otro en una secuencia. Por lo tanto debemos buscar a An.

Hay un valor inicial que es 5000, una ganancia y una perdida de clientes, que podemos representar así:

An = 5000 + Ganancia - Perdida

Inicial = A0 = 5000

Ganancia = 100, pero cómo sucede cada año, sería: 100 * n

Perdida = 10% de los miembros actuales, si al principio son 5000 por lo tanto A0 * 0.10, pero en este primer caso es que es A0, pero en general serían An-1

reemplazando:

An = A0 + 100 * n - (An-1) * 0.10

7 0
4 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
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nirvana33 [79]
An angle of 200 degrees in an obtuse angle.
6 0
3 years ago
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