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3 years ago

One algebra question please!

Mathematics

1 answer:

jeyben [28]3 years ago

3 0

Hello from MrBillDoesMath!

Answer: 3x^3 + 6 x^2 - 12 x (the third choice)

Discussion:

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

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

3x^3 + 6 x^2 - 12 x

which is the third choice

Thank you,

MrB

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Candy selling at six dollars per pound will be mixed with candy selling at nine dollars per pound how many pounds of more expens

podryga [215]

The 15 lbs at $7=$105

So multiple 5x9 as a test and check results. 105-45=60

Here you can see that 60 is a factor of 6 by 10 so you got lucky and got the answer of 10lbs at $6 & 5lbs at $9=15 lbs at $7 average

3 0

3 years ago

For each rectangle below, write a linear equation that represents the area y of the rectangle. Solve this system of two linear e

Hatshy [7]

(5,24)

The two areas are the same.

To find the area, we multiply the side lengths. Y=area

Rectangle 1: side lengths 4 and (x+1)

y=4(x+1)= 4x+4

Rectangle 2: side lengths 3 and (2x-2)

y=3(2x-2)= 6x-6

Since the two areas are same, we can conclude that

4x+4=6x-6

Now, simplify.

-2x=-10, x=5

Since x is 5, we can plug it into the equations to find y.

Option 1 with rectangle 1: y=4(5)+4, y=24

Option 2 with rectangle 2: y=6(5)-5, y=24

I graphed the linear equation on desmos.

5 0

2 years ago

M(4,2) is the midpoint of RS. The coordinates of S are (6, 1). what are the coordinates of R?

nirvana33 [79]

Remark
You are using the midpoint formula. Instead of finding the midpoint, you are looking for one of the points, so you have to rearrange the formula a little bit.

Givens
Midpoint (4,2)
One endpoint (6,1)

Object
Find the other endpoint.

Formula
m(x,y) = (x1 + x2)/2, (y1 + y2)/2)

Solution
Find the x value
4 = (6 + x2)/2 Multiply both sides by 2
4*2 = 6 + x2 Subtract 6 from both sides.
8 - 6 = x2
x2 = 2

Find the y value
2 = (1 + y2)/2 Multiply by 2
4 = 1 + y2 Subtract 1 from both sides.
4 - 1 = y2
y2 = 3

Conclusion
R(x,y) = (2,3)

8 0

3 years ago

The third-degree Taylor polynomial about x = 0 of In(1 - x) is

gizmo_the_mogwai [7]

Answer:

$\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation

Calculus

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

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

MacLaurin/Taylor Polynomials

Approximating Transcendental and Elementary functions
MacLaurin Polynomial: $\displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n$
Taylor Polynomial: $\displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n$

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

Step 1: Define

Identify

f(x) = ln(1 - x)

Center: x = 0

n = 3

Step 2: Differentiate

[Function] 1st Derivative: $\displaystyle f'(x) = \frac{1}{x - 1}$
[Function] 2nd Derivative: $\displaystyle f''(x) = \frac{-1}{(x - 1)^2}$
[Function] 3rd Derivative: $\displaystyle f'''(x) = \frac{2}{(x - 1)^3}$

Step 3: Evaluate Functions

Substitute in center x [Function]: $\displaystyle f(0) = ln(1 - 0)$
Simplify: $\displaystyle f(0) = 0$
Substitute in center x [1st Derivative]: $\displaystyle f'(0) = \frac{1}{0 - 1}$
Simplify: $\displaystyle f'(0) = -1$
Substitute in center x [2nd Derivative]: $\displaystyle f''(0) = \frac{-1}{(0 - 1)^2}$
Simplify: $\displaystyle f''(0) = -1$
Substitute in center x [3rd Derivative]: $\displaystyle f'''(0) = \frac{2}{(0 - 1)^3}$
Simplify: $\displaystyle f'''(0) = -2$

Step 4: Write Taylor Polynomial

Substitute in derivative function values [MacLaurin Polynomial]: $\displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3$
Simplify: $\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}$

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0

3 years ago

Corina used standard algorithm. and multiplied 62×22 and got the product 1,042 explain why Corina's answer is not reasonable

Sliva [168]

62
×22
=124
+1240
=1364
her answer is not responsible because how can 124+1240=1,042

3 0

4 years ago