What side is this a isosceles a scalene or equilateral

I'll give brainlest if u can answer this question 1+1=?

yKpoI14uk [10]

Answer:

11

Step-by-step explanation:

LOL

3 0

3 years ago

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What is 60% of 115 show your work

tangare [24]

115 x (.60) = 69
nwkdcw

8 0

3 years ago

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4(2x+3)+4(3x+2) simplify and expand answer

timama [110]

Answer:

20x + 20

Step-by-step explanation:

4 ( 2x + 3 ) + 4 ( 3x + 2 )

Take 4 common from both,

4 ( 2x + 3 + 3x + 2 )

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

= 4 ( 5x + 5 )

= 4 ( 5x ) + 4 ( 5 )

= 20x + 20

4 0

3 years ago

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Is the function represented by the table non-linear?

dexar [7]

Answer:

Nope!

Step-by-step explanation:

It's linear, because it has a continuous pattern. Look!

x values: 6, 7, 8, 9

y values: 4, 2, 0, -2

X values: We add 1 each time

Y values: We subtract 2 each time

I hope it helps!

$*RosalindCordelia*$

8 0

3 years ago

\int\limits^0_∞ cos{x} \, dx

JulijaS [17]

Answer:

$\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty)$

General Formulas and Concepts:

Pre-Calculus

Unit Circle
Trig Graphs

Calculus

Limits
Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$
Integrals
Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$
Trig Integration
Improper Integrals

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^0_\infty {cos(x)} \, dx$

Step 2: Integrate

[Improper Integral] Rewrite: $\displaystyle \lim_{a \to \infty} \int\limits^0_a {cos(x)} \, dx$
[Integral] Trig Integration: $\displaystyle \lim_{a \to \infty} sin(x) \bigg| \limits^0_a$
[Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle \lim_{a \to \infty} sin(0) - sin(a)$
Evaluate trig: $\displaystyle \lim_{a \to \infty} -sin(a)$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle -sin(\infty)$

Since we are dealing with infinity of functions, we can do a numerous amount of things:

Since -sin(x) is a shift from the parent graph sin(x), we can say that -sin(∞) = sin(∞) since sin(x) is an oscillating graph. The values of -sin(x) already have values in sin(x).
Since sin(x) is an oscillating graph, we can also say that the integral actually equates to undefined, since it will never reach 1 certain value.

∴ $\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty) \ or \ \text{unde}\text{fined}$

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

Unit: Improper Integrals

Book: College Calculus 10e

7 0

3 years ago