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Dima020 [189]
3 years ago
8

Represent each of the following variables in the appropriate letters.Next determine wether the variable has a fixed value or a c

hanged value.
a)The score Sarah got in a Mathematics test.
b)The distance between the Zen house and the school.
c)Temperature on the summit of Mount Everest in a day.​
Mathematics
1 answer:
Alex777 [14]3 years ago
6 0

Step-by-step explanation:

a) Let 'm' represents score in a Mathematics test. The variable has a fixed value because the marks Sarah obtained in her test will not change

b) Let 'd' represent the distance between the Zen house and the school. The variable has a fixed value because by using the same route, the distance will always remain constant.

c) Let 't' represent the temperature on the summit of Mount Everest. The variable has a changed value because the temperature may alter depending on the time of day and also climatic conditions which will affect the temperature.

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You want one of the values in the ratio to be one for a unit rate. First, write out the ratio. It is 70 miles/2 hours. It's better not to include decimal numbers when you write a ratio when you can, so we're going to make the hours the unit rate. To make 2 -> 1 hour, divide both sides of the ratio by 2. 70/2 = 35. 
The unit rate is 35 mi./hr.
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Two friends are training for the marathon. Shaneese runs 100 miles per week, give or take 5 miles. Denasia runs 20 miles per wee
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Answer:

give or take 5 miles.

Step-by-step explanation:

im not shor but that is true about the both of them

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4. Find the standard from of the equation of a hyperbola whose foci are (-1,2), (5,2) and its vertices are end points of the dia
Ad libitum [116K]

The <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

<h3>How to find the standard equation of a hyperbola</h3>

In this problem we must determine the equation of the hyperbola in its <em>standard</em> form from the coordinates of the foci and a <em>general</em> equation of a circle. Based on the location of the foci, we see that the axis of symmetry of the hyperbola is parallel to the x-axis. Besides, the center of the hyperbola is the midpoint of the line segment with the foci as endpoints:

(h, k) = 0.5 · (- 1, 2) + 0.5 · (5, 2)

(h, k) = (2, 2)

To determine whether it is possible that the vertices are endpoints of the diameter of the circle, we proceed to modify the <em>general</em> equation of the circle into its <em>standard</em> form.

If the vertices of the hyperbola are endpoints of the diameter of the circle, then the center of the circle must be the midpoint of the line segment. By algebra we find that:

x² + y² - 4 · x - 4 · y + 4= 0​

(x² - 4 · x + 4) + (y² - 4 · y + 4) = 4

(x - 2)² + (y - 2)² = 2²

The center of the circle is the midpoint of the line segment. Now we proceed to determine the vertices of the hyperbola:

V₁(x, y) = (0, 2), V₂(x, y) = (4, 2)

And the distance from the center to any of the vertices is 2 (<em>semi-major</em> distance, a) and the semi-minor distance is:

b = √(c² - a²)

b = √(3² - 2²)

b = √5

Therefore, the <em>standard</em> form of the equation of the hyperbola that satisfies all conditions is (x - 2)²/4 - (y - 2)²/5 = 1 .

To learn more on hyperbolae: brainly.com/question/27799190

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5 0
1 year ago
(b) dy/dx = (x - y+ 1)^2
Elanso [62]

Substitute v(x)=x-y(x)+1, so that

\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}

Then the resulting ODE in v(x) is separable, with

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On the left, we can split into partial fractions:

\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx

Integrating both sides gives

\dfrac{\ln|1-v|+\ln|1+v|}2=x+C

\dfrac12\ln|1-v^2|=x+C

1-v^2=e^{2x+C}

v=\pm\sqrt{1-Ce^{2x}}

Now solve for y(x):

x-y+1=\pm\sqrt{1-Ce^{2x}}

\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}

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