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3 years ago

What is the equation of a hyperbola with a = 6 and c = 8? Assume that the transverse axis is horizontal.

1 answer:

3 0

The standard equation for a hyperbola with a horizontal transverse axis is

(x-h)^2/a^2 - (y-k)^2/b^2 = 1 . The center is at (h, k) . The distance between the vertices is 2a . The distance between the foci is 2c and c^2 = a^2 + b^2. Therefore,

8^2 = 6^2 + b^2
b = 2√7
Assuming it is located at the origin.

(x-h)^2/a^2 - (y-k)^2/b^2 = 1
(x)^2/6^2 - (y)^2/(2√7)^2 = 1
x^2/36 - y^2/28 = 1

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What is the reason why the earth rotates

qaws [65]

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3 years ago

The measure of angle RTP is 50 degrees. Find the measure of angle TPS. 140 65 40

mr Goodwill [35]

Answer:

Step-by-step explanation:

90 degree angle

90 - 50 = 40

4 0

3 years ago

Solve the quadratic equation give your answer to 2 decimal places : 3x^2+x-5=0

Dima020 [189]

Given:

The quadratic equation is:

$3x^2+x-5=0$

To find:

The solution for the given equation rounded to 2 decimal places.

Solution:

Quadratic formula: If a quadratic equation is $ax^2+bx+c=0$ , then:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

We have,

$3x^2+x-5=0$

Here, $a=3,b=1,c=-5$ . Using the quadratic formula, we get

$x=\dfrac{-1\pm \sqrt{1^2-4(3)(-5)}}{2(3)}$

$x=\dfrac{-1\pm \sqrt{1+60}}{6}$

$x=\dfrac{-1\pm \sqrt{61}}{6}$

$x=\dfrac{-1\pm 7.81025}{6}$

Now,

$x=\dfrac{-1+7.81025}{6}$

$x=1.13504167$

$x\approx 1.14$

And

$x=\dfrac{-1-7.81025}{6}$

$x=-1.468375$

$x\approx -1.47$

Therefore, the required solutions are 1.14 and -1.47.

3 0

3 years ago

PLEASE ANSWER AND SHOW WORKING OUT FOR ME PLEASE.

Lelu [443]

7932 x 56 = 444192 and since it’s asking you to multiply by .56 and 793.2, just move the decimal over to the left three places in your answer since there’s three digits after the decimals in your factors. so you get 444.192

7 0

4 years ago

Read 2 more answers

Find the sum of the given vectors and illustrate geometrically. [0,1,2],[0,0,-3]

Nitella [24]

Answer:

The sum of the given vectors is $[0,1,2]+[0,0,-3]=[0,1,-1]$

Step-by-step explanation:

For the given vectors (which are R³ vectors), the sum is simply the sum of each coordinate, if a general vector is written as

$[x,y,z]$

the sum of two vector will be in each coordinate at a time.

To illustrate geometrically the resulting vector in the space xyz

$[0,1,-1]$

we can say that the first coordinate is on the x-axis, the sencond on the y-axis, and the third one on the z-axis, so the illustration will be a vector starting from the center of coordinates, and ending in the coordinates 0 of the x-axis, 1 of the y-axis, and -1 of the z-axis. Or, in a plane yz (where x=0), a vector from the origin to the point 1 in y-axis, and -1 in z-axis.

5 0

3 years ago