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3 years ago

Hat is the equation of the quadratic function with a vertex at (2,–25) and an x-intercept at (7,0)?

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

y = x² - 4x - 21

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (2, - 25), thus

y = a(x - 2)² - 25

To find a substitute (7, 0) into the equation

0 = a(7 - 2)² - 25 = a(5)² - 25 = 25a - 25 ( add 25 from both sides )

25a = 25 ( divide both sides by 25

a = 1, thus

y = (x - 2)² - 25 ← in vertex form

Expand and simplify

y = x² - 4x + 4 - 25

y = x² - 4x - 21 ← in standard form

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The total cost (in dollars) for a company to manufacture and sell x items per week is C = 40 x + 640 , whereas the revenue broug

ikadub [295]

Answer:

The company must sell 60 or 70 items to obtain a weekly profit of 200.

Step-by-step explanation:

The profit is the difference between the revenue and the cost of a given task, therefore:

$\text{profit} = R - C\\\text{profit} = 66*x - 0.2*x^2 - (40*x + 640)\\\text{profit} = 66*x - 0.2*x^2 - 40*x - 640\\\text{profit} = - 0.2*x^2 + 26*x - 640$

To have a profit of 200, we need to sell:

$-0.2*x^2 + 26*x - 640 = 200\\-0.2*x^2 + 26*x -840 = 0\text{ } *\frac{-1}{0.2}\\x^2 -130 + -4200 = 0\\x_{1,2} = \frac{-(-130) \pm \sqrt{(-130)^2 - 4*1*(-4200)}}{2*1}\\x_{1,2} = \frac{130 \pm \sqrt{16900 + 16800}}{2}\\x_{1,2} = \frac{130 \pm \sqrt{100}}{2}\\x_{1,2} = \frac{130 \pm 10}{2}\\x_{1} = \frac{130 + 10}{2} = \frac{140}{2} = 70\\ x_{2} = \frac{130 - 10}{2} = \frac{120}{2} = 60$

The company must sell 60 or 70 items to obtain a weekly profit of 200.

3 0

3 years ago

Please help me with precalculus??<br>linear and angular speed

Anit [1.1K]

13)

there are 2π radians in 1 revolution, and there are 60 seconds in 1 minute, so keeping that in mind, then,

$\bf \cfrac{4\underline{\pi} }{5~\underline{s}}\cdot \cfrac{rev}{2\underline{\pi} }\cdot \cfrac{60~\underline{s}}{min}\implies \cfrac{4\cdot 60~rev}{5\cdot 2~min}\implies \cfrac{240~rev}{10~min}\implies 24\frac{rev}{min}$

14)

$\bf \textit{linear velocity}\\\\
v=rw\quad 
\begin{cases}
r=radius\\
w=angular~speed\\
----------\\
v=32\frac{m}{sec}\\
w=100\frac{rev}{min}
\end{cases}\\\\
-------------------------------\\\\
\textit{let's convert \underline{w} to }\frac{radians}{sec}$

$\bf \cfrac{100~\underline{rev}}{\underline{min}}\cdot \cfrac{2\pi }{\underline{rev}}\cdot \cfrac{\underline{min}}{60~sec}\implies \cfrac{100\cdot 2\pi }{60~sec}\implies \cfrac{10\pi }{3~sec}\implies \cfrac{10\pi }{3}\frac{radians}{sec}\\\\
-------------------------------\\\\
v=rw\implies \cfrac{v}{w}=r\implies \cfrac{\frac{30~m}{sec}}{\frac{10\pi }{3~sec}}\implies r=\cfrac{30~m}{\underline{sec}}\cdot \cfrac{3~\underline{sec}}{10\pi }
\\\\\\
r=\cfrac{90}{10\pi }m$

15)

what is the radians per seconds "w" in revolutions per minute? just another conversion like in 13)

$\bf \cfrac{\underline{\pi} }{3~\underline{sec}}\cdot \cfrac{rev}{2\underline{\pi }}\cdot \cfrac{60~\underline{sec}}{min}\implies \cfrac{60 ~rev}{3\cdot 2 ~min}\implies \cfrac{60 ~rev}{6 ~min}\implies 10\frac{rev}{min}$

4 0

3 years ago

Four times the square of a non-zero number is equal to twelve times the number.

Whitepunk [10]

4x^2 = 12x
4x^2 - 12x = 0
4x(x - 3) = 0

4x = 0
x = 0...not this one

x - 3 = 0
x = 3 <== ur number

4 0

3 years ago

What is the area of a triangle with a height of 13 cm and a base of a 10 cm?

Alecsey [184]

Area of a triangle is equal to 1/2 b(h) so since you already have the height and base, all you need to do is input it back into the equation:

$\frac{1}{2}$ (10)(13)

Multiply 10 & 13 to equal

$\frac{1}{2}$ (130)

Then take half of 130 to get

A = 65 $cm^{2}$

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3 years ago

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Korolek [52]

Y=4 or -4
Explanation: trust me bro

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3 years ago

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