All categories

2 years ago

EXTRA PTS name the relationship between the pair of angles

Mathematics

2 answers:

mars1129 [50]2 years ago

7 0

21 21 21 21 21 21 21 21 21 21 21 21 21

Harrizon [31]2 years ago

4 0

I’m assuming there is a picture you are missing.

You might be interested in

Besides 18 and 1, what is one factor of 18?

Andreyy89

Answer:

6, 2

Step-by-step explanation:

8 0

2 years ago

Solve: –3x – 24 ≤ –36 what is x equal to?

Kay [80]

Answer:

X <= 4

Step-by-step explanation:

-3x - 24<= -36

Make -3x the subject, by carrying -24 to the other side

Note: Moving -24 to the other side changes to + 24

-3x <= -36 + 24

-3x <= - 12

Minus cancels minus

3x <= 12

Divide both sides by 3,to get the value of x

3x/3 <= 12/3

x <= 12/3

x <= 4

Therefore, x equals to 4

6 0

3 years ago

Read 2 more answers

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

s344n2d4d5 [400]

From the first equation
6x - y = 1
6x - y - 1 = 0
6x - 1 = y

Substituting we get
4x - 3(6x - 1) = -11

5 0

2 years ago

2. What is the overall purpose of photosynthesis?

mafiozo [28]

Answer:

plants getting there food is the basic use of photosynthesis

Step-by-step explanation:

6 0

2 years ago

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

*EXTRA PTS* name the relationship between the pair of angles

EXTRA PTS name the relationship between the pair of angles