Xy=10
x+y=29
subtract x from both sides for second equation
y=-x-29
subsitute in second equaiton
x(-x-29)=10
-x^2-29x=10
add x^2+29x to both sides
x^2+29x+10=0
quadratic formula
if you have
ax^2+bx+c=0,
x=
so
x^2+29x+10=0
a=1
b=29
c=10
x=
x=
x=
x=
x=
or
those are the 2 numbers
aprox=-28.65 and -0.349
Answer:
The equation i.e. used to denote the population after x years is:
P(x) = 490(1 + 0.200 to the power of x
Step-by-step explanation:
This problem could be modeled with the help of a exponential function.
The exponential function is given by:
P(x) = ab to the power of x
where a is the initial value.
and b=1+r where r is the rate of increase or decrease.
Here the initial population of the animals are given by: 490
i.e. a=490
Also, the rate of increase is: 20%
i.e. r=20%
i.e. r=0.20
Hence, the population function i.e. the population of the animals after x years is:
P(x) = 490(1 + 0.200 to the power of x
5k/2G is the answer to the problem.
Answer:
a) 7/4 b) 15/2
Step-by-step explanation:
a) Scale Factor = Image / Object
Take the side of Triangle ABC that's BC Why? Because it corresponds with the Image Triangle ADE's DE
= 21/12 = 7/4
b) 7/4 = 10+ x / 10
Cross Multiply
= 4 ( 10 + x ) = 70
= 40 + 4x = 70
= 4x = 70-40
= 4x = 30
= 4x/4 = 30/4
x = 15/2
