20 % of 2 is equal to A. 20 B. 4 C. 0.4 D. 0.04

Two boats depart from a port located at (–8, 1) in a coordinate system measured in kilometers and travel in a positive x-directi

miss Akunina [59]

Answer:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

Step-by-step explanation:

1st boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a$

Equation:

$y=ax^2 -2ax+c$

The y-coordinate of the vertex:

$y_v=a\cdot 1^2-2a\cdot 1+c\Rightarrow a-2a+c=10\\ \\c-a=10$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1$

Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

7 0

3 years ago

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It cost Faith $9.10 to send 91 text messages. How many text messages did she send if she spent $17.60?

Komok [63]

Answer:

um

Step-by-step explanation:

176 10cent per 1 message

4 0

2 years ago

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A ball is thrown into the air from an apartment balcony and falls to the ground. The height of the ball, h metres, t seconds aft

andriy [413]

Answer:

$1+2\sqrt{2}$ seconds

Step-by-step explanation:

When the ball hits the ground, its height will obviously be 0. Therefore, you can set up the equation the following way:

$0=-5t^2+10t+35 \\\\-5(t^2-2t-7)=0$

Plugging this into the quadratic equation, you get:

$t=\dfrac{2\pm\sqrt{4+28}}{2}= \\\\\dfrac{2\pm4\sqrt{2}}{2}= \\\\1\pm2\sqrt{2}$

Since the time must be positive, it takes the ball $1+2\sqrt{2}$ seconds to hit the ground, or around 3.828. Hope this helps!

7 0

3 years ago

Please help me quick ASAP!!

Crank

A hope this helps
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7 0

3 years ago

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A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times.

tatyana61 [14]

Answer:

(a) $g(n)=0.05\cdot 2^n$

(b) $g^{-1}(n)=\log_{2}20g(n)$

(c)43 times

Step-by-step explanation:

Part A

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

$g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n$

Part B

$g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)$

Part C

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

$g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43$

You must fold the paper 43 times to make the folded paper have a thickness that is the same as the distance from the earth to the moon.

3 0

3 years ago