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2 years ago

SUBJECT: Algebra LESSON: Multiplying Polynomials (x^3 + 2x − 3)(x^4 − 3x^2 + x)

Mathematics

1 answer:

NNADVOKAT [17]2 years ago

7 0

Answer:

Step-by-step explanation:

(x^3 + 2x − 3)(x^4 − 3x^2 + x)

Multiply each value of 2nd bracket with 1st bracket:

=x^4(x^3 + 2x − 3) - 3x^2(x^3 + 2x − 3) +x(x^3 + 2x − 3)

=x^7+2x^5-3x^4-3x^5-6x^3+9x^2+x^4+2x^2-3x

Now combine the terms with same power:

=x^7-x^5-2x^4-6x^3+11x^2-3x

You can also take the common from the expression:

x(x^6-x^4-2x^3-6x^2+11x-3)....

The product of (x^3 + 2x − 3)(x^4 − 3x^2 + x) is x(x^6-x^4-2x^3-6x^2+11x-3)....

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What goes into 8 and 18

Lunna [17]

Step-by-step explanation:

common factor 8 and 18 is 2.

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2 years ago

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Find the equation of the line that is parallel to the line x + 5y = 10 and passes through the point (1,3).

Nataliya [291]

Answer:

Step-by-step explanation:

for two lines to be parallel they MUST have the same slope but have different Y intercepts ( Y axis crossing values)

parallel lines will never intersect each other

first I use a graphing calcuator to solve the problem

I like to use y intercept form better, but you don't have use it

I will try to solve it directly without going through the y intercept equation

x + 5y = 10 in y intercept form y = mx + b is

5y = -x + 10 divide both sides by 5

y = -x/5 + 10/5

y = -x/5 + 2

Now I need to find a parallel line that passes through the x =1 and y = 3 point recall this line will have the same slope = m = -1/5

and a different Y axis crossing point 'b' that we don't know

y = -x/5 + b y = 3 when x = 1 so slove for 'b'

3 = -1/5 + b

3 = -0.2 + b add -0.2 to both sides

3.2 = b

y = -x/5 + 3.2 this is answer in y intercept form

if you multiply both sides by 5

5y = -x + 16 add x to both sides

x + 5y = 16 answer in standard form

THE EASIER WAY TO SOLVE THE PROBLEM IS ......

x + 5y = 10 find a line parallel that passes through x = 1, y = 3

x + 5y = Z Z is an unknown value plug in x and y and solve for Z

1 + 5(3) = Z

Z = 16 so the parallel line in standard form is

x + 5y = 16

same as before, my y intercept method was correct, but not worth the effort

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3 years ago

There are 12 students who wish to enroll in a particular course. there are only five seats left in the classroom. how many diffe

siniylev [52]

792 is the answer

(12*11*10*9*8)/5!=792

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

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3 years ago

Select the statement that best justifies the conclusion based on the given information.

CaHeK987 [17]

9514 1404 393

Answer:

c. If a plane contains a line, it contains the points on the line.

Step-by-step explanation:

The only statement relating a point on a line to the plane containing the line is the one shown above.

_____

Additional comment

Identifying true statements is a reasonable strategy for many multiple-choice questions. Another strategy that can be employed is finding the one true statement that is relevant to the question being asked.

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2 years ago