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4 years ago

Need help with this one keep getting it wrong

Mathematics

1 answer:

Fittoniya [83]4 years ago

6 0

The correct option is D. In fact, you can evaluate this function at every possible input x.

All this function does is:

Take a number x
Add 6
If the result is negative switch the sign, otherwise keep it as it is

As you can see, you can perform these operations regardless of the input x.

You might be interested in

17 gallons of water are used to completely fill 4 fish tanks. If each tank holds the same amount of water how many gallons will

Kay [80]

Each tank holds 4.25 gallons! 17/4 is 4.25 :) hope this helps!

6 0

3 years ago

If anyone could help that would be great! The topic is calculus, and substitution + integrals. I need help with the ones circled

Alex_Xolod [135]

Answer:

$\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Trig Derivatives

Integration

Integrals
Definite Integrals
Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration

Logarithmic Integration

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 1 + cos(x)$
[u] Differentiate [Trig Derivative]: $\displaystyle du = -sin(x) \ dx$
[Bounds of Integration] Change: $\displaystyle [\frac{1}{2}, 1]$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{-sin \ x}{1 + cos \ x}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -\int\limits^{1}_{\frac{1}{2}} {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -(ln|u|) \bigg| \limits^{1}_{\frac{1}{2}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^{\frac{-\pi}{2}}_{\frac{-2 \pi}{3}} {\frac{sin \ x}{1 + cos \ x}} \, dx = -ln(2)$

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

Unit: Integration

Book: College Calculus 10e

8 0

3 years ago

I need help it’s a quiz

belka [17]

Answer:

answer is

Step-by-step explanation:

Ok first you dial 911 and study for the test you un smart kid then dig a whole jump in it and you will be in a fire place after that you spend rest of etrnity there good luck.

5 0

3 years ago

Read 2 more answers

What is the percent decrease from 60 to 39?

xxMikexx [17]

$That\ is\ a\ decrease\ of\ \boxed{35}\ percent.$