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german
3 years ago
5

Which describes the graph of y= (x+ 6)²+ 1? A. Vertex at (-6,-1) B. Vertex at (-6, 1) C. Vertex at (6, 1) D. Vertex at (6,-1)​

Mathematics
1 answer:
mamaluj [8]3 years ago
3 0

Answer:

B

Step-by-step explanation:

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Prove identity tan 3x−tan 2x−tanx = tan 3xtan 2xtanx
NeX [460]
Tan ( a + b ) = [ tan a + tan b ] / [ 1 - (tan a)*(tan b) ];

let be a = 2x and b = x;

=> tan 3x = [ tan 2x + tan x ] / [ 1 - (tan 2x)*(tan x) ] => (tan 3x)*[ 1 - (tan 2x)*(tan x) ] =
 
tan 2x + tan x => tan3x -  tan 3xtan 2xtanx = tan 2x + tan x => <span> tan 3x−tan 2x−tanx = tan 3xtan 2xtanx.</span><span />
7 0
3 years ago
(cotx+cscx)/(sinx+tanx)
Butoxors [25]

Answer:   \bold{\dfrac{cot(x)}{sin(x)}}

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)

\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}

6 0
3 years ago
Pls help me for brainliest answer answer needs to be correct
ira [324]

Answer:

a) 17, 29, 41, 53, 65....

b) 12n + 5

3 0
3 years ago
1. In an auditorium, there are 21 seats in the first row and 26 seats in the second row. The number of seats in a row continues
kondor19780726 [428]
1. Let s_n be the number of seats in the n-th row. The number seats in the n-th row relative to the number of seats in the (n-1)-th row is given by the recursive rule

s_n=s_{n-1}+5


Since s_1=21, we have

s_2=s_1+5
s_3=s_2+5=s_1+2\cdot5
s_4=s_3+5=s_1+3\cdot5
\cdots
s_n=s_{n-1}+5=\cdots=s_1+(n-1)\cdot5

So the explicit rule for the sequence s_n is

s_n=21+5(n-1)\implies s_n=5n+16

In the 15th row, the number of seats is


s_{15}=5(15)+16=91

2. Let p_n be the amount of profit in the n-th year. If the profits increase by 6% each year, we would have

p_2=p_1+0.06p_1=1.06p_1
p_3=1.06p_2=1.06^2p_1
p_4=1.06p_3=1.06^3p_1
\cdots
p_n=1.06p_{n-1}=\cdots=1.06^{n-1}p_1

with p_1=40,000.

The second part of the question is somewhat vague - are we supposed to find the profits in the 20th year alone? the total profits in the first 20 years? I'll assume the first case, in which we would have a profit of


p_{20}=1.06^{19}\cdot40,000\approx121,024

3. Now let p_n denote the number of pushups done in the n-th week. Since 3\cdot4=12, 12\cdot4=48, and 48\cdot4=192, it looks like we can expect the number of pushups to quadruple per week. So,

p_n=4p_{n-1}

starting with p_1=3.

We can apply the same reason as in (2) to find the explicit rule for the sequence, which you'd find to be

p_n=4^{n-1}p_1\implies p_n=4^{n-1}\cdot3
6 0
3 years ago
What is C(-2,6),D(10,-8)
IceJOKER [234]
They are coordinates
5 0
3 years ago
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