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uranmaximum [27]
3 years ago
15

Square root of -1? PLEASE HELP

Mathematics
1 answer:
kolezko [41]3 years ago
7 0

My understanding it would be letter ( i )  because it is imaginary.  

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In relation t a functions? Is the inverse of relation t function
frozen [14]

Answer:

Relation t is a function. The inverse of relation t is not a function ⇒ 3rd

Step-by-step explanation:

* Lets explain how to solve the problem

- A relation is a set of inputs and outputs, and a function is a relation

 with one output for each input

- Ex: T = {(1 , 2) , (3 , 5) , (-4 , 0)} is a function because every input has

  only one output

- To find the inverse of a function we switched the input and the

  out put

- The inverse of a function may not always be a function

* Lets solve the problem

∵ The relation t is the set of ordered pairs of x and y

  x = 0     2      4    6

  y = -8    -7    -4    -4

∵ x = 0 has only y = -8

∵ x = 2 has only y = -7

∵ x = 4 has only y = -4

∵ x = 6 has only y = -4

∴ Every value of x has only one value of y

∴ Relation t is a function

* Lets find its inverse by switching x and y

∵ The inverse function of t is :

   x = -8     -7     -4    -4

   y =  0      2      4     6

∵ x = -8 has only y = 0

∵ x = -7 has only y = 2

∵ x = -4 has y = 4 and y = 6

∴ Not every value of x has only one value of y

∴ The inverse of relation t is not a function

* Relation t is a function. The inverse of relation t is not a function

4 0
3 years ago
This is due today, please help!!! I will give brainliest.
WINSTONCH [101]

Answer:

Step 1, all the exponents are increased by 4

Step-by-step explanation:

The first incorrect step occurred in Step 1, where all the exponents were increased by 4.

This is mathematically incorrect due to exponential rules. When distributing exponents inside parentheses, we have to multiply the existing exponents inside the parentheses by the exponent outside the parentheses.

For example, (x²)³ is not x²⁺³, but rather, x⁽²⁾⁽³⁾.

We multiply the exponents instead of adding them together.

Therefore, the correct Step 1 should multiply all the variables' exponents by 4.

Steps 2 and 3 are correct since we do add the exponents when multiplying exponents with the same base, and we do subtract exponents with the same base when dividing.  

6 0
3 years ago
Solve for y 16y^2-25=0
Pepsi [2]

Answer:

\large\boxed{x=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}

Step-by-step explanation:

16y^2-25=0\\\\METHOD\ 1:\\\text{use}\ a^2-b^2=(a-b)(a+b)\\\\16=4^2\ \text{and}\ 25=5^2\ \text{therefore we have}\\\\4^2y^2-5^2=0\\\\(4y)^2-5^2=0\\\\(4y-5)(4y+5)+0\iff4y-5=0\ \vee\ 4y+5=0\\\\4y-5=0\qquad\text{add 5 to both sides}\\4y=5\qquad\text{divide both sides by 4}\\\boxed{y=\dfrac{5}{4}}\\\\4y+5=0\qquad\text{subtract 5 from both sides}\\4y=-5\qquad\text{divide both sides by 4}\\\boxed{x=-\dfrac{5}{4}}

METHOD\ 2:\\\\16y^2-25=0\qquad\text{add 25 to both sides}\\\\16y^2=25\qquad\text{divide both sides by 16}\\\\y^2=\dfrac{25}{16}\to y=\pm\sqrt{\dfrac{25}{26}}\\\\y=-\dfrac{\sqrt{25}}{\sqrt{16}}\ \vee\ x=\dfrac{\sqrt{25}}{\sqrt{16}}\\\\\boxed{y=-\dfrac{5}{4}\ \vee\ x=\dfrac{5}{4}}

7 0
3 years ago
Read 2 more answers
Solve for x: y=-3x+6
vazorg [7]
The answer is : x= -y/3+2
8 0
3 years ago
Read 2 more answers
Can someone help me with this​
SashulF [63]

Answer:

\angle NRQ =85 \degree

Given:

\angle RPQ = 45 \degree \\  \angle PQR = (6x + 4)\degree \\  \angle NRQ = (15x - 5)\degree

Step-by-step explanation:

Property used: <em>An exterior angle of a triangle is equal to the sum of the opposite int</em><em>e</em><em>rior angles.</em>

<em>=  >  \angle RPQ +  \angle PQR =  \angle NRQ \\  \\  =  > 45 \degree + (6x + 4)\degree = (15x - 5)\degree \\  \\  =  > 45\degree + 6x\degree + 4\degree = 15x\degree - 5\degree \\  \\  =  > 49\degree  + 5 \degree= 15x\degree - 6x\degree \\  \\  =  > 54\degree = 9x\degree \\  \\  =  > 9x\degree = 54\degree \\  \\  =  > x\degree =  (\frac{54}{9} )\degree \\  \\  =  > x\degree = 6\degree \\  \\ Putting \: value \: of \: x \: in \:  \angle NRQ  \\   \\  =  > \angle NRQ = (15x - 5)\degree \\  \\  =  >  \angle NRQ = (15 \times 6 - 5) \degree \\  \\  =  >  \angle NRQ =85 \degree</em>

4 0
3 years ago
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