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2 years ago

(40 pts)can you factor the difference of square?perfect square trinomials?

Mathematics

1 answer:

stira [4]2 years ago

7 0

Answer:

Step-by-step explanation:

The difference of two squares may be represented by the formula: a^2-b^2,

which can be factored as (a+b)(a-b)

A perfect square trinomial may be represented by the formula: a^(2)-2ab+b^2 or a^(2)+2ab+b^2, depending on the sign of b

if b is negative: use the formula a^(2)-2ab+b^2, which can be factored as (a-b)*(a-b) or (a-b)^(2)

if b is positive: use the formula a^(2)+2ab+b^2, which can be factored as (a+b)*(a+b) or (a+b)^(2)

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How many terms are in the following expression:

Airida [17]

Answer:4

Step-by-step explanation:

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2 years ago

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Help me pls I've been waiting for an hour to get an answer I'm willing to sacrifice my points ...

Sveta_85 [38]

Answer:

I think it's probably 8$ Edit: 7.5$ i did the math wrong

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2 years ago

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac

Fed [463]

Answer:

$\theta=k\pi, \hspace{3}k\in Z\\\\ or\\\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z\\$

Step-by-step explanation:

Factor constant terms:

$-2(-4sin(2\theta)+sin(\theta))=0$

Divide both sides by -2:

$sin(\theta)-4sin(2 \theta)=0$

Expand trigonometric functions using the fact:

$sin(2 \theta) =2 sin(\theta) cos(\theta)$

So:

$sin(\theta) -8sin(\theta)cos(\theta)=0$

Factor sin(x) and constant terms and multiply both sides by -1:

$sin(\theta) (8cos(\theta)-1)=0$

Split into two equations:

$(1)=8cos(\theta)-1=0\\\\(2)=sin(\theta)=0$

For (1)

Add 1 to both sides and divide both sides by 8:

$cos(\theta)=\frac{1}{8}$

Take the inverse cosine of both sides:

$\theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z$

For (2)

Simply take the inverse sine of both sides

$\theta = k \pi, \hspace{3}k\in Z$

Therefore, the solutions are given by:

$\theta=k\pi, \hspace{3}k\in Z\\\\ or\\ \theta =2\pi k \pm arccos(\frac{1 }{8} ), \hspace{3}k\in Z\\$

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3 years ago

What is the vertex of the parabola given by the equation y x^2+ 4?

max2010maxim [7]

Format of Quadratic Equation: y = ax2 + bx + c

Given Quadratic Equation: y = 2x2 - 3x + 3

Coefficient Variable Values: a = 2 and b = -3 and c = 3

Axis of Symmetry: x = -b/2a = -(-3)/2(2) so answer is x = 3/4

Vertex: x value is axis of symmetry (3/4) and y value is calculated substituting 3/4 for x in original equation: y = 2(3/4)2 - 3(3/4) + 3 = 2(9/16) - 9/4 + 3 = 9/8 - 9/4 + 3 = 9/8 - 18/8 + 24/8 = 15/8,
so answer is (3/4,15/8)

x intercepts (solve using quadratic formula): x = (-b plus or minus sqrt(b2 - 4ac)/2a, so plugging in coefficient values for a and b and c, we get x = [-(-3) plus or minus sqrt((-3)2 - 4(2)(-3)]/2(2), which results in x = (3 + sqrt(33))/4 or (3 - sqrt(33))/4 and answers to nearest tenth are x = (3 + 5.7) / 4 = 2.2
or x = (3 - 5.7) / 4 = -0.7

y intercept is calculated by substituting zero for x into original equation: y = 2x2 - 3x + 3, so y-intercept is 3.

Domain is range is from calculated negative x intercept (-0.7) to calculated positive x intercept (2.2) and is written as (-0.7,2.2)

Range is from calculated y-intercept to positive infinity, since parabola opens up due to positive x2 coefficient value, so range is written as (3,positive infinity). Note: infinity symbol is sideways 8.

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3 years ago

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Simplify: 3x2 - 27<br> Helppppp

oksian1 [2.3K]

3x^2- 27 … you divide the whole thing by 3 to get x^2- 9 and that’s the furthest you can simplify

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2 years ago