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3 years ago

14

The average rainfall in the Florida Keys for January through March is 1.7 inches per month. On average, what is the total rainfa

ll for the three months?

1 answer:

OleMash [197]3 years ago

8 0

I have a feeling that it is 5.1 inches

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In the diagram, a circle centered at the origin, a right triangle, and the Pythagorean theorem are used to derive the equation o

ehidna [41]

Answer:

(x – h)2 + (y – k)2 = r2

Step-by-step explanation:

If the center of the circle were moved from the origin to the point (h, k) and point P at (x, y) remains on the edge of the circle the equation of the new circle

(x – h)2 + (y – k)2 = r2

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3 years ago

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Gina ate 3/4 of an apple pie. greg ate 1/8 of the same pie. how much of the Apple pie was left?

miskamm [114]

There would be 1/8 of the pie left,
you have to add 3/4 and 1/8 to find the total eaten and then subtract it by 8/8 (the total) which would give you your answer.

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3 years ago

Paul and five friends each sold the same number of tickets. They sold 54 tickets altogether. How many tickets did each person se

zloy xaker [14]

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2 years ago

PLs help...............

alexdok [17]

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Step-by-step explanation:

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What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

VladimirAG [237]

Answer:

the positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

$\frac{a}{b}$ to be the slope in the equation $y = \pm \frac{10}{2} (y-11)+2$

$\frac{a}{b}$ = $\frac{10}{2}$

$\frac{a}{b}$ = 5

Thus, the positive slope of the asymptote = 5

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3 years ago