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3 years ago

Select the conic section that represents the equation. x2 - y2 = 16

Mathematics

1 answer:

Mariana [72]3 years ago

8 0

$x^2-y^2=16$ represents a hyperbola.

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Which graph shows a proportional relationship?

Verdich [7]

Answer: The only graph that shows a proportional relationship is the line that crosses the origin point (0,0).

Explanation

The other graphs are linear functions but not not proportional relationships.

The general form of a proportional relationship is y = kx, where k is the proportionality constant. So, for x = 0 you will always obtain y = 0.

The general form of a linear relationshio is y = kx + b, being b the y-intercept, so if the y-intercept is not 0, it is not a proportional relationship. That is what happens with the other three graphs.

6 0

3 years ago

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What is the quotient of 6,135 and 15

zavuch27 [327]

Answer:

409

Step-by-step explanation:

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3 years ago

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Six different second-year medical students at Bellevue Hospital measured the blood pressure of the same person. The systolic re

Yanka [14]

Answer:

$Range=14$

$\sigma^2 =32.4$

$\sigma = 5 .7$

The standard deviation will remain unchanged.

Step-by-step explanation:

Given

$Data: 136, 129, 141, 139, 138, 127$

Solving (a): The range

This is calculated as:

$Range = Highest - Least$

Where:

$Highest = 141; Least = 127$

So:

$Range=141-127$

$Range=14$

Solving (b): The variance

First, we calculate the mean

$\bar x = \frac{1}{n} \sum x$

$\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)$

$\bar x = \frac{1}{6} *810$

$\bar x = 135$

The variance is calculated as:

$\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2$

So, we have:

$\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]$

$\sigma^2 =\frac{1}{5}*[162]$

$\sigma^2 =32.4$

Solving (c): The standard deviation

This is calculated as:

$\sigma = \sqrt {\sigma^2 }$

$\sigma = \sqrt {32.4}$

$\sigma = 5 .7$ --- approximately

Solving (d): With the stated condition, the standard deviation will remain unchanged.

5 0

3 years ago

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constru

Viktor [21]

Answer:

1) $S = 2\cdot w\cdot l - 8\cdot x^{2}$ , 2) The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$ , 3) $S = 176\,in^{2}$ , 4) $x \approx 4.528\,in$ , 5) $S = 164.830\,in^{2}$

Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

3) The surface area when a 1'' x 1'' square is cut out is:

$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

$x \approx 4.528\,in$

5) The equation of the box volume is:

$V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x$

$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

$V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^{2} + 4\cdot x^{3}$

$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

The first derivative of the function is:

$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

The critical points are determined by equalizing the derivative to zero:

$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

$x_{1} \approx 4.952\,in$

$x_{2}\approx 1.548\,in$

The second derivative is found afterwards:

$V'' = 24\cdot x - 78\,in$

After evaluating each critical point, it follows that $x_{1}$ is an absolute minimum and $x_{2}$ is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is: