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seropon [69]
3 years ago
5

(3a3 − 5b3) + _____ a3 + _____b3 = (a3 + b3).

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
7 0
3a^3-5b^3+xa^3+yb^3=1a^3+1b^3
replace a^3 with z
replace b^3 with v (to make it easier to see)
3z-5v+xz+yv=z+v
3z+xz+yv-5v=1z+v

the terms match them
3z+xz=1z
divide both sides by z
3+x=1
minus 3
x=-2

yv-5v=1v
dividie both sides by v
y-5=1
addd 5 to both sides
y=6



(3a^3-5b^3)+ <u>-2</u> a^3 +<u>6</u> b^3=(a^3+b^3)

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Graph A:  y = (1/2)x - 2

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5 0
2 years ago
Five over twenty five÷ 100
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20

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3 0
3 years ago
Which of the following is a solution to 2cos2x − cos x − 1 = 0?
Pavel [41]

Answer:

Option A is correct.

Solution for the given equation is, x = 0^{\circ}

Step-by-step explanation:

Given that : 2\cos^2x -\cos x -1 =0

Let \cos x =y

then our equation become;

2y^2-y-1= 0           .....[1]

A quadratic equation is of the form:

ax^2+bx+c =0.....[2] where a, b and c are coefficient and the solution is given by;

x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}

Simplify:

y = \frac{ 1 \pm \sqrt{1+8}}{4}

or

y = \frac{ 1 \pm \sqrt{9}}{4}

y = \frac{ 1 \pm 3}{4}

or

y = \frac{1+3}{4} and y = \frac{1 -3}{4}

Simplify:

y = 1 and y = -\frac{1}{2}

Substitute y = cos x we have;

\cos x = 1

⇒x = 0^{\circ}

and

\cos x = -\frac{1}{2}

⇒ x = 120^{\circ} \text{and} x = 240^{\circ}

The solution set:  \{0^{\circ}, 120^{\circ} , 240^{\circ}\}

Therefore, the solution for the given equation  2\cos^2x -\cos x -1 =0 is, 0^{\circ}





8 0
3 years ago
Read 2 more answers
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