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2 years ago

Plzzzzzzzzzzzzzz i really need someone to help me!!! ASAP!!

Mathematics

2 answers:

Pavlova-9 [17]2 years ago

5 0

The second answer I believe. Sorry if I am wrong.

yKpoI14uk [10]2 years ago

3 0

Your answer is D.
If it were a reflection, it would be the same, just flipped as if you flipped a piece of paper over by it's long edge/side. A is out.
B would make it be an entirely different direction
C would just move it a little to the right.
D would flip it completely, as if you flipped a piece of paper over by it's short side.

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Fill in the bubble next to the correct answer

alexdok [17]

424 divided by four= 106 the buuble is correct

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2 years ago

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Suppose that the relation G is defined as follows. G={(4, -4), (-6,0), (-6, -7), (-1, 4)}Give the domain and range of G. Write y

SIZIF [17.4K]

Let's draw a Venn diagram of the given relation:

The domain is the set of all first elements of ordered pairs (x-coordinates) and the range is the set of all second elements (y -coordinates).

Therefore, from the picture above, the answer is:

$\begin{gathered} \text{DOMAIN: }\mleft\lbrace-6,-1,4\mright\rbrace \\ \text{RANGE:}\mleft\lbrace-4,0,4,7\mright\rbrace \end{gathered}$

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1 year ago

Find the solutions to x⁴-64=0 and the x-intercepts of the graph of y=x⁴-64.

Setler [38]

Answer:

1. $x_1=2\sqrt{2}\\x_2=-2\sqrt{2}\\x_3=i2\sqrt{2}\\x_4=-i2\sqrt{2}$

2. $A=(2\sqrt{2},0)$ and $B=(-2\sqrt{2},0)$

Step-by-step explanation:

We have the expression $x^4-64=0$ to find the solutions we have to clear $x$ . Then,

$x^4-64=0$ Add 64 in both sides of the equation.

$x^4-64+64=64\\x^4=64$

Now we have to re-write the equation:

$u=x^2$ ⇒ $u^2=(x^2)^2\\u^2=x^4$

Then,

$u^2=64$

Apply square root on both members

$\sqrt{u^2}=\sqrt{64}\\u=8 , u=-8$

Now substitute back $u=x^2$

1. $x^2=8$

2. $x^2=-8$

Solving for x:

1. $x=+\sqrt{8}=\sqrt{2^3}=\sqrt{2^2}.\sqrt{2}\\x=2\sqrt{2}\\or\\x=-\sqrt{8}\\x=-2\sqrt{2}$

2. $x=+\sqrt{-8}=\sqrt{(-1).8} =\sqrt{-1}\sqrt{8} \\x=i\sqrt{8}\\x=i.2\sqrt{2} \\or\\x=-\sqrt{-8}\\x=-i.2\sqrt{2}$

Then the solutions for $x^4-64=0$ are:

$x_1=2\sqrt{2}\\x_2=-2\sqrt{2}\\x_3=i2\sqrt{2}\\x_4=-i2\sqrt{2}$

To find the x intercepts of the graph $y=x^4-64$ we have to replace with y=0.

This means: $x^4-64=0$

We already found the solutions for the expression, but we have to consider only the real solutions.

$x_1=2\sqrt{2}\\x_2=-2\sqrt{2}$

Because in a graph we can't have imaginary solutions.

Then the x intercepts of the graph are:

$A=(2\sqrt{2},0)$ and $B=(-2\sqrt{2},0)$

The graph of the function is:

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3 years ago

6x – 3y + 2z = -30<br> 6x + 5y=-4<br> -6z= -18

Assoli18 [71]

Answer:

x = 251/765, y = 698/85, z = 7/3

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2 years ago

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Korvikt [17]

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2 years ago