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3 years ago

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Mathematics

2 answers:

GREYUIT [131]3 years ago

8 0

Answer:

$a_1=4,a_2=6,a_3=8,a_4=10$

Step-by-step explanation:

We are given that a graph

We can see that point on the graph

(1,4),(2,6),(3,8),(4,10).

We have to find the arithmetic sequence

Sequence :It is that function which domain is natural set and range is real number .

Arithmetic sequence : It is that sequence in which difference between two consecutive terms are equal.

Sequence like as

$f(1),f(2),f(3),.. f(n)$

Or $a_1,a_2,a_3,...,a_n$

We have

$a_1=4,a_2=6,a_3=8,a_4=10$

masha68 [24]3 years ago

6 0

The answer is
First term: 2
Second Term: -2
Third: -6
Fourth: -10.

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Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

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

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

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