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shutvik [7]
1 year ago
8

Hii! would anyone mind helping me out?

Mathematics
1 answer:
Eduardwww [97]1 year ago
5 0

Answer:

(4, 4)

Step-by-step explanation:

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

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

Match the values in this hyperbola to those of the standard form. The variable h represents the x-offset from the origin, k represents the y-offset from origin, a.

a = 1

b = 3

k = 4

h = 4

Thus,

(x-4)^2 - \frac{(y-4)^2}{9} = 1

The center of a hyperbola follows the form of (h, k). Substitute in the values of h and k.

= (4, 4)

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Division property of equality, because you have to divide 8x and 80 by 10

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3 years ago
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2m - 3 - 5= -m + 10 please help me
valina [46]

\qquad\qquad\huge\underline{\boxed{\sf Answer☂}}

Let's solve ~

\qquad \sf  \dashrightarrow \: 2m - 3 - 5 =  - m + 10

\qquad \sf  \dashrightarrow \: 2m + m = 10 + 3 + 5

\qquad \sf  \dashrightarrow \: 3m = 18

\qquad \sf  \dashrightarrow \: m = 18 \div 3

\qquad \sf  \dashrightarrow \: m = 6

Hence, value of m is 6

3 0
2 years ago
Pi/24 is the solution for 4cos2 (4x) - 3 = 0<br><br> True or false
inessss [21]

Answer:

True.

Step-by-step explanation:

Given,

4cos2 (4x) - 3 = 0

Now, 4π/24 = π/6

Or, cos π/6 = √3/2

Or, cos^2 π/6 = 3/4

Or, 4 cos^2 π/6 = 3

Now, Left Hand Side= 4cos2 (4x) - 3

                                  = 3-3 =0 = Right Hand Side (Proved)

Here, Left Hand Side= Right Hanad Side. So, 4cos2 (4x) - 3=0 is true.

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2 years ago
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1.<br> Which equation is true when m = 12?
kobusy [5.1K]

Answer:

are there any more details? cant solve anything from what you have provided

Step-by-step explanation:

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2 years ago
50PTS!<br> Describe the vertical asymptotes and holes for the graph of y = x-5/x^2 -1
Diano4ka-milaya [45]
Ok

first of all, for q(x)/p(x)
if the degree of q(x) is less than the degree of p(x),then the horizontal assemtote is 0

then simplify
any factors you factored out is now a hole, remember them

to find the vertical assemtotes of a function, set the SIMPLIFIED denomenator equal to 0 and solve

so

y=(x-5)/(x^2-1)
q(x)<p(x)
horizontal assemtote is y=0

no factors to simplify so no holes

set denomenator to 0 to find vertical assemtote
x^2-1=0
(x-1)(x+1)=0
x-1=0
x=1

x+1=0
x=-1

the horizontal assemtotes are x=1 and -1
3 0
2 years ago
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