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aniked [119]
3 years ago
12

What are the foci of the hyperbola with equation 5y^2-4x^2=20?

Mathematics
2 answers:
shusha [124]3 years ago
5 0
The foci of the hyperbola with equation 5y^2-4x^2=20 will be given as follows:
divide each term by 20
(5y^2)/20-(4x^2)/20=20/20
simplifying gives us:
y^2/4-x^2/5=1
This follows the standard form of the hyperbola
(y-k)²/a²-(x-h)²/b²=1
thus
a=2, b=√5 , k=0, h=0
Next we find c, the distance from the center to a focus.
√(a²+b²)
=√(2²+(√5)²)
=√(4+5)
=√9
=3
the focus of the hyperbola is found using formula:
(h.h+k)
substituting our values we get:
(0,3)
The second focus of the hyperbola can be found by subtracting c from k
(h,k-c)
substituting our values we obtain:
(0,-3)

Thus we have two foci
(0,3) and (0,-3)
ExtremeBDS [4]3 years ago
3 0
(0 <span>±</span> 3) is the answer to your question? Hope this helps
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If f(x) = 5x + 40, what is f(x) when x = -5?<br> 0 -9<br> O-8<br> 07<br> 0 15
gogolik [260]

Good morning ☕️

______

Answer:

15

___________________

Step-by-step explanation:

f(x) = 5x + 40

then

f(-5) = 5(-5) + 40

= -25 + 40

= 40-25

= 15.

:)

3 0
3 years ago
a mailing tube in the shape of cylinder has a height of 24 inches and a radius of 3 inches. what is the volume, in cubic inches,
TiliK225 [7]

ANSWER

The volume of the mailing tube is 678.6 square inches to the nearest tenth.

EXPLANATION

The volume of a cylinder is calculated using the formula;

V=\pi \:  {r}^{2} h

Since the mailing tube is in the form of a cylinder, we use this formula to find its volume.

We substitute h=24 and r=3 to obtain,

V=\pi \times {3}^{2} \times 24

V=216\pi  {in}^{2}

V=678.6{in}^{2}

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3 0
3 years ago
What are the zeros of the function f(x) = x2 + 5x + 5 written in simplest radical form?
Pavel [41]

\boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Explanation:</h2>

Using the quadratic formula:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \\ Here: \\ \\ f(x) = x^2 + 5x + 5 \\ \\ \\ So: \\ \\ a=1 \\ \\ b=5 \\ \\ c=5 \\ \\ \\ x=\frac{-5 \pm \sqrt{5^2-4(1)(5)}}{2(1)} \\ \\ x=\frac{-5 \pm \sqrt{25-20}}{2} \\ \\ x=\frac{-5 \pm \sqrt{5}}{2} \\ \\ \\ Two \ solutions: \\ \\ \boxed{x_{1}=\frac{-5 + \sqrt{5}}{2}} \\ \\ \\ \boxed{x_{2}=\frac{-5 - \sqrt{5}}{2}}

<h2>Learn more:</h2>

Quadratic functions: brainly.com/question/12164750

#LearnWithBrainly

8 0
3 years ago
The product of three integers is-3 Determine all of the possible values for the three factors
miskamm [114]
3 * 1 * -1 because 3 times one is 3 and that times a negative turns it into a negative. ( positive * negative = negative)
8 0
3 years ago
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