Answer:
Step-by-step explanation:
We first have to write the equation for the sequence, then finding the first five terms will be easy. Follow the formatting:
and we are given enough info to fill in:
and
and
or in linear format:
where n is the position of the number in the sequence. We already know the first term is -35.
The second term:
so
and
The third term:
and
so
and we could go on like this forever, but the nice thing about this is when we know the difference all we have to do is add it to each number to get to the next number.
That means that the fourth term will be -27 + 4 which is -23.
The fifth term then will be -23 + 4 which is -19. You can check yourself by filling in a 5 for n in the equation and solving:
and
so
Answer:
L(x,y) = (2,-8,0) + (0,-1,1)*t
Step-by-step explanation:
for the planes
x + y + z = -6 and y + z = -8
the intersection can be found subtracting the equation of the planes
x + y + z - ( y + z ) = -6 - (-8)
x= 2
therefore
x=2
z=z
y= -8 - z
using z as parameter t and the point (2,-8,0) as reference point , then
x= 2
y= -8 - t
z= 0 + t
another way of writing it is
L(x,y) = (2,-8,0) + (0,-1,1)*t
The angles of a triangle always equal 180.
You do 180 - 80 - 45. X = 45.
1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that 
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:
Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹
The solution for this problem is:
The population is 500 times bigger since 8000/24 = 500. The population after t days is computed by:P(t) = P₀·4^(t/49)
Solve for t: 8000 = 8·4^(t/49) 1000 = 4^(t/49) log₄(1000) = t/49t = 49log₄(1000) ≅ 244 days
