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gogolik [260]
1 year ago
5

Factorize: x²+2+1/x² ​

Mathematics
1 answer:
andrew-mc [135]1 year ago
8 0

x^{2}+\frac{1}{x^{2}}+2=\boxed{\left(x+\frac{1}{x} \right)^{2}}

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Please help me with precal
Doss [256]

Consider the function f(x)=\sqrt{x}. This function has:

  • the domain x\in [0,\infty);
  • the range y\in [0,\infty).

If the domain of unknown function is [a,\infty), then x\ge a or x-a\ge 0. This means that you have \sqrt{x-a} as a part of needed function.

If the range of unknown function is [b,\infty), then y\ge b. This means that you have to translate function b units up and then the expression of the function is

y=\sqrt{x-a}+b.

Answer: correct choice is B.

6 0
2 years ago
PLA HELP QUESTION IN PHOTO
asambeis [7]

Answer:

Step-by-step explanation:

its the first one

7 0
2 years ago
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If cosθ=2/3 find cscθ using identities. This is in quad. 1.
laiz [17]

Step-by-step explanation:

  • cos θ = 2/3

\quad \twoheadrightarrow\sf {cos \; \theta = \dfrac{Base}{Hypotenuse} } \\

Hence, base = 2 units and hypotenuse = 3 units.

\quad \twoheadrightarrow\sf { H^2 = B^2 + P^2} \\

\quad \twoheadrightarrow\sf { P^2 = H^2 - B^2} \\

\quad \twoheadrightarrow\sf { P^2 = (3)^2 - (2)^2} \\

\quad \twoheadrightarrow\sf { P^2 = 9- 4} \\

\quad \twoheadrightarrow\bf { P = \sqrt{5}} \\

Now, we know that :

\quad \twoheadrightarrow\sf {cosec \; \theta = \dfrac{Hypotenuse}{Perpendicular} } \\

\quad \twoheadrightarrow\bf{cosec \; \theta = \dfrac{3}{\sqrt{5}} } \\

Therefore, the required answer is 3/√5.

4 0
2 years ago
What is the vertex form, f(x) = a(x - h)2-k, for a parabola that passes through the point (2,8) and has (1,0) as its
Novay_Z [31]

Answer:

Step-by-step explanation:

hello :

f(x) = a(x - h)²-k   ....vertex form  when the vertex is : (h,k)

h=1  and  k=0   so : f(x) = a(x-1)²

a parabola that passes through the point (2,8) : f(2)=8

a(2-1)² =8

a= 8

f(x) = 8(x-1)²=  8(x²-2x+1)

f(x) 8x²-16x+8   .....standard form

3 0
3 years ago
How to solve for X: X-B=A
mamaluj [8]

Answer:

Step-by-step explanation:

(x−a)(x−b)=x2−(a+b)x+ab

Now, this with the third bracket.

(x2−(a+b)x+ab)(x−c)=x3−(a+b+c)x2+(ac+bc+ab)x−abc

But there’s another way to do this, which is easier. Assume the given expression is equal to 0, then, we can form a cubic equation as

x3−(sum−of−roots)x2+(product−of−roots−taken−two−at−a−time)x−(product−of−roots) , which is essentially what we got above.

5 0
3 years ago
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