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4 years ago

6(5x-3) distributive property gcf

2 answers:

jeka944 years ago

8 0

6(5x-3)

Multiply each inside number by the outside number
6•5x= 30x
6•-3= -18

Put them back into their equation
30x-18

omeli [17]4 years ago

3 0

Answer:

Step-by-step explanation:30x-18

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The green object below,is best defined as a ___ named____.

notka56 [123]

Answer:

ray AC

Step-by-step explanation:

hope thsi helpes!

6 0

3 years ago

Read 2 more answers

Question 2 (2 points)

hoa [83]

Answer:

4) The limit does not exist.

General Formulas and Concepts:

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$
Left-Side Limit: $\displaystyle \lim_{x \to c^-} f(x)$

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

Step-by-step explanation:

*Note:

For a limit to exist, the right-side and left-side limits must be equal to each other.

Step 1: Define

Identify

$\displaystyle f(x) = \left\{\begin{array}{ccc}5 - x ,\ x < 5\\8 ,\ x = 5\\x + 3 ,\ x > 5\end{array}$

Step 2: Find Left-Side Limit

Substitute in function [Left-Side Limit]: $\displaystyle \lim_{x \to 5^-} 5 - x$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^-} 5 - x = 5- 5 = 0$

Step 2: Find Left-Side Limit

Substitute in function [Right-Side Limit]: $\displaystyle \lim_{x \to 5^+} x + 3$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \lim_{x \to 5^+} x + 3 = 5 + 3 = 8$

∴ since $\displaystyle \lim_{x \to c^+} f(x) \neq \lim_{x \to c^-} f(x)$ , $\displaystyle \lim_{x \to 5} f(x) = \text{DNE}$

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

Unit: Limits

5 0

3 years ago

How do you do this question?

lukranit [14]

Answer:

0.001591

Step-by-step explanation:

The power series for arctan(x) is:

arctan(x) = ∑ (-1)ⁿ x²ⁿ⁺¹ / (2n + 1)

Substituting 5x:

arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺¹ / (2n + 1)

Multiply both sides by x:

x arctan(5x) = ∑ (-1)ⁿ x (5x)²ⁿ⁺¹ / (2n + 1)

Simplify:

x arctan(5x) = ∑ (-1)ⁿ (5x) (5x)²ⁿ⁺¹ / (10n + 5)

x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺² / (10n + 5)

Multiply top and bottom by 5:

x arctan(5x) = ∑ (-1)ⁿ 5 (5x)²ⁿ⁺² / (50n + 25)

Integrate:

∫ x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))

Evaluate between x = 0.1 and x = 0:

∫₀⁰¹ x arctan(5x) = [∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))] |₀⁰¹

∫₀⁰¹ x arctan(5x) = ∑ (-1)ⁿ (0.5)²ⁿ⁺³ / ((50n + 25) (2n + 3))

This is an alternating series. We can approximate it with Alternating Series Estimation.

bₙ₊₁ ≥ ε

(0.5)²⁽ⁿ⁺¹⁾⁺³ / ((50(n+1) + 25) (2(n+1) + 3)) ≥ 0.000001

(0.5)²ⁿ⁺⁵ / ((50n + 75) (2n + 5)) ≥ 0.000001

n ≥ 3

So the approximation is the sum of the terms from n=0 to n=3.

(-1)⁰ (0.5)²⁽⁰⁾⁺³ / ((50(0) + 25) (2(0) + 3))

+ (-1)¹ (0.5)²⁽¹⁾⁺³ / ((50(1) + 25) (2(1) + 3))

+ (-1)² (0.5)²⁽²⁾⁺³ / ((50(2) + 25) (2(2) + 3))

+ (-1)³ (0.5)²⁽³⁾⁺³ / ((50(3) + 25) (2(3) + 3))

= 0.0016667 − 0.0000833 + 0.0000089 − 0.0000012

= 0.001591

7 0

3 years ago

Simplify 2x^2 - x^2, not solving for c

MArishka [77]

I hope this helps you 2x^2-x^2=x^2

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4 years ago

Which expression is equal to 5/2?

vlabodo [156]

my answer was D I think it is the one

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3 years ago