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makvit [3.9K]
3 years ago
15

(07.03 MC) An equation is shown below: 9(3x − 16) + 15 = 6x − 24 Part A: Write the steps you will use to solve the equation, and

explain each step. (6 points) Part B: What value of x makes the equation true? (4 points)
Mathematics
1 answer:
Alik [6]3 years ago
6 0

Answer: See steps and explanation below. The value 5 makes the equation true.

9(3x - 16) + 15 = 6x - 24                Distributive Property

27x - 144 + 15 = 6x - 24                Combine like terms

27x - 129 = 6x - 24                       Subtract 6x from both sides

21x - 129 = -24                              Add 129 to both sides

21x = 105                                      Divide both sides by 21

x = 5                                             Answer!

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Encuentre el radio y el área de círculos cuya circunferencias tienen las siguientes medidas
Vadim26 [7]

Answer:

Los radios de los círculos son r_{1} \approx 9.995, r_{2} \approx 2.013, r_{3} \approx 7.592, respectivamente.

Las áreas de los círculos son A_{1} \approx 313.845, A_{2} \approx 12.730, A_{3} \approx 181.077, respectivamente.

Step-by-step explanation:

La circunferencia (s) se calcula mediante la siguiente fórmula:

s = 2\pi\cdot r (1)

Donde r es el radio del círculo.

Una vez hallado el radio, se determina el área de la figura geométrica (A) mediante la siguiente fórmula:

A = \pi\cdot r^{2} (2)

Si conocemos que las circunferencias son s_{1} = 62.8, s_{2} = 12.65 y s_{3} = 47.7, respectivamente:

1) Radios de los círculos

r_{1} = \frac{s_{1}}{2\pi}, r_{2} = \frac{s_{2}}{2\pi}, r_{3} = \frac{s_{3}}{2\pi}

r_{1} \approx 9.995, r_{2} \approx 2.013, r_{3} \approx 7.592

2) Áreas de los círculos

A_{1} = \pi\cdot r_{1}^{2}, A_{2} = \pi\cdot r_{2}^{2}, A_{3} = \pi\cdot r_{3}^{2}

A_{1} \approx 313.845, A_{2} \approx 12.730, A_{3} \approx 181.077

6 0
3 years ago
Solve for x. <br><br> −1/3x≤−6
Olin [163]
-1/3x</= -6
/-1/3      /-1/3
x</= +2
x>/=+2     
The sign changes because you are dividing by a -
7 0
3 years ago
What are the vertex, axis of symmetry, maximum or minimum value, and range of y = 3x^2 + 6x − 1?
Bess [88]
The vertex is when x=-b/2a, which is also the line of symmetry. 
in this case, b=6, a=3, so x=-6/6=-1
when x=-1. y=3-6-1=-4

another way to do it is to write the equation into vertex form by making a square:
3x²+6x-1
3(x²+2x-1/3)
3(x²+2x+1-1-1/3))
3(x²+2x+1-4/3))
notice the bold part makes a square:
3(x+1)²-4

Either way, 
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the axis of symmetry is x=-1
the coefficient of x² is 3, a positive number, so this parabola opens upward. the function has a minimum value of y=-4
the range is all real number larger than -4
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Find the zeros of the function.<br> f(x) = 2x2 – 2x – 12
Amiraneli [1.4K]

Answer:

Step-by-step explanation:

2x^2-2x-12

This has solutions of x=3 and x= -2

3 0
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The value of quarters in the bowl on week 1 was &5
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