Answer:
1) -184
2) 13
-3(
)+4
Step-by-step explanation:
For problem number 1, Let's start by substituting all the x's for -3
So the equation will look like
p(-3)=
-
+7(-3)-10
Now, lets simplify by solving the exponents
First we'll do 
which equals -27
Now we'll do 
which equals 9
So now our equation looks like
p(-3)=4(-27)-5(9)+7(-3)-10
Now, we are going to multiply all these numbers
First 4(-27) which equals -108
Now -5(9) which equals -45
And finally 7(-3) which equals -21
Now our equation looks like
p(-3)= -108-45-21-10
Now we can solve
-108-45-21-10=-184
Answer:
3
Step-by-step explanation:
All factorials 5 and above are evenly divisible by 15, so have no remainder. Thus, you are interested in ...
mod(1! +2! +3! +4!, 15) = mod(1 +2 +6 +24, 15)
= mod(33, 15) = 3
The remainder is 3.
What grade are you?
i can hopple u
Answer:
it is C
Step-by-step explanation:
Answer:
(e) csc x − cot x − ln(1 + cos x) + C
(c) 0
Step-by-step explanation:
(e) ∫ (1 + sin x) / (1 + cos x) dx
Split the integral.
∫ 1 / (1 + cos x) dx + ∫ sin x / (1 + cos x) dx
Multiply top and bottom of first integral by the conjugate, 1 − cos x.
∫ (1 − cos x) / (1 − cos²x) dx + ∫ sin x / (1 + cos x) dx
Pythagorean identity.
∫ (1 − cos x) / (sin²x) dx + ∫ sin x / (1 + cos x) dx
Divide.
∫ (csc²x − cot x csc x) dx + ∫ sin x / (1 + cos x) dx
Integrate.
csc x − cot x − ln(1 + cos x) + C
(c) ∫₋₇⁷ erf(x) dx
= ∫₋₇⁰ erf(x) dx + ∫₀⁷ erf(x) dx
The error function is odd (erf(-x) = -erf(x)), so:
= -∫₀⁷ erf(x) dx + ∫₀⁷ erf(x) dx
= 0
