Mistake in line one
He forgot to factor the 2 out of -8x
Pakal triangle is hard to do with long things, but it is easy up to like 5 degree
we look at the row for 5th degree (6th row)
the sequence is
1,5,10,10,5,1
that is the coeficients
for (a+b)^5 that is
1a^5b^0+5a^4b^1+10a^3b^2+10a^2b^3+5a^1b^4+1a^0b^5
see the exponents each time add to 5
so
1x^5(-5)^0+5x^4(-5)^1+10x^3(-5)^2+10x^2(-5)^3+5x^1(-5)^4+1x^0(-5)^5=
x^5-25x^4+250x^3-1250
Answer: y = 16.5
Step-by-step explanation:
We would apply Pythagoras theorem which is expressed as
Hypotenuse² = opposite side² + adjacent side²
Considering triangle BCD,
BD² = 14² - 8²
BD² ° 196 - 64 = 132
BD = √132
Considering triangle ABD,
AB² = BD² + y²
AB² = (√132)² + y²
AB² = 132 + y²- - - - - - - - - - 1
Considering triangle ABC,
(8 + y)² = AB² + BC²
(8 + y)² = AB² + 14²
(8 + y)² = AB² + 196
AB² = (8 + y)² - 196- - - - - - - -2
Substituting equation 1 into equation 2, it becomes
132 + y² = (8 + y)² - 196
y² - (8 + y)² = - 196 - 132
y² - (64 + 16y + y²) = - 328
y² - 64 - 16y - y2 = - 328
- 16y = - 328 + 64
- 16y = - 264
y = - 264/- 16
y = 16.5
The solution is A
x=2 y=-1
the solution is where the two lines intersect
Answer:
7
Step-by-step explanation:
add 7 to 3x
Then: 4x-3x=1x which gives:
1x=7
7/1=7
