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3 years ago

Please please please help!! ASAP

Mathematics

1 answer:

crimeas [40]3 years ago

5 0

I prefer to use a midpoint formula that deals with both coordinates at once. For points A and B, their midpoint (M) is

... M = (A + B)/2

We can solve this for B by multiplying by 2 and subtracing A.

... 2M = A + B

... 2M - A = B

For your midpoint M(0, -10) and your endpoint A(-4, -5), the other endpoint will be

... B = 2(0, -10) - (-4, -5) = (2·0 -(-4), 2·(-10) -(-5))

... = (4, -15)

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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:

1. $D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}$ .

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

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

6 0

3 years ago

Y=-2x+5 if x=4 what is the answer

barxatty [35]

Answer:

y=-3

Step-by-step explanation:

y=-2x+5

y=-2(4)+5=-8+5=-3

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andrew11 [14]

Answer:

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What is the y-intercept of the function f(x) = –2/9x + 1/3?

marishachu [46]

f(x) = –2/9x + 1/3
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answer
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