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Ganezh [65]
3 years ago
15

Reflections. plsss helpp will mark brainliest :) if you don't know don't put anything at all. :)

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
3 0

Answer:

(5, 2)

Step-by-step explanation:

If we're reflecting across the y-axis, the y value of the point stays the same, because we're going from left to right in this case.  The x point, on the other hand, gets inverted (- to + and + to -).  -5 becomes 5 and 2 stays the same.

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Dean is saving up change so he can go to the arcade with 100 quarters. If he has 363 pennies, 200 nickels, and 23 dimes. How muc
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D. $9.07
explanation: 363 pennies is equal to $3.36, 200 nickels is equal to $10.00, and 23 dimes is equal to $2.30. all of this change added up is equal to $15.93. since 100 quarters are worth $25, you would take 25-15.93. this equals D.$9.07
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What is the meaning of the term percentage?​
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What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.
lapo4ka [179]

Consider the equation x^6+6x^3+5=0.

First, you can use the substitution t=x^3, then x^6=(x^3)^2=t^2 and equation becomes t^2+6t+5=0. This equation is quadratic, so

D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1.

Then you can factor this equation:

(t+5)(t+1)=0.

Use the made substitution again:

(x^3+5)(x^3+1)=0.

You have in each brackets the expression like a^3+b^3 that is equal to (a+b)(a^2-ab+b^2). Thus,

x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)

and the equation is

(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0.

Here x_1=-\sqrt[3]{5} , x_2=-1 and you can sheck whether quadratic trinomials have real roots:

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2. D_2=(-1)^2-4\cdot 1=1-4=-3.

This means that quadratic trinomials don't have real roots.

Answer: x_1=-\sqrt[3]{5} , x_2=-1

If you need complex roots, then

x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2}   ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}.

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