Answer:
they are all prime numbers
Answer:
2x + y - 6 = 0
OR 2x + y = 6
Step-by-step explanation:
First write the equation given in the problem:
y = -2x + 6 This is in slope-intercept form (y = mx + b).
Standard form is written Ax + By + C = 0. When C is a negative number, you might also see it as Ax + By = -C.
The main difference between the two forms in that slope-intercept form isolates the 'y' whereas standard form equates to 0. Don't confuse the 'b' in standard from with the 'B' in slope-intercept form.
To convert from slope-intercept form to standard form, <u>move everything over to the side with 'y'</u>. When you move something, you do its reverse operation to the whole equation. (The reverse of addition is subtraction, the reverse of multiplication is division.)
y = -2x + 6 Do the reverse operations for -2x and +6
y + 2x - 6 = -2x + 2x + 6 - 6 Add 2x and subtract 6 on both sides
y + 2x - 6 = 0 Right side cancels out to be '0'.
2x + y - 6 = 0 Rewrite with the 'x' in front of the 'y'
Here you can see the new equation and what each variable in Ax + By + C = 0 is.
A = 2
B = 1 When a number is not written with the variable, it is 1.
C = -6
Some teachers ask it to be rewritten as Ax + By = -C when 'C' is a negative number.
2x + y = 6
Answer:
Step-by-step explanation:
1.
cot x sec⁴ x = cot x+2 tan x +tan³x
L.H.S = cot x sec⁴x
=cot x (sec²x)²
=cot x (1+tan²x)² [ ∵ sec²x=1+tan²x]
= cot x(1+ 2 tan²x +tan⁴x)
=cot x+ 2 cot x tan²x+cot x tan⁴x
=cot x +2 tan x + tan³x [ ∵cot x tan x
=1]
=R.H.S
2.
(sin x)(tan x cos x - cot x cos x)=1-2 cos²x
L.H.S =(sin x)(tan x cos x - cot x cos x)
= sin x tan x cos x - sin x cot x cos x

= sin²x -cos²x
=1-cos²x-cos²x
=1-2 cos²x
=R.H.S
3.
1+ sec²x sin²x =sec²x
L.H.S =1+ sec²x sin²x
=
[
]
=1+tan²x ![[\frac{\textrm{sin x}}{\textrm{cos x}} = \textrm{tan x}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Ctextrm%7Bsin%20x%7D%7D%7B%5Ctextrm%7Bcos%20x%7D%7D%20%3D%20%5Ctextrm%7Btan%20x%7D%5D)
=sec²x
=R.H.S
4.

L.H.S=



= 2 csc x
= R.H.S
5.
-tan²x + sec²x=1
L.H.S=-tan²x + sec²x
= sec²x-tan²x
=


=1
The answer to the question is <span>70.6179775281. which also can be converted to 71.</span>
Answer:
1
Step-by-step explanation:
hope this help:) I use slop formula