Answer:
(-2.2, -1.6), (3, 1)
Step-by-step explanation:
You don't have to go far to find the equations. They are right there in your problem statement. Perhaps you want to find the solutions to the equations.
Use the first equation to write an expression for x, then substitute that into the second equation:
x = 2y +1
y^2 -3(2y+1)(y) +8 = 0
-5y^2 -3y +8 = 0
-(5y +8)(y -1) = 0
y = -8/5 or y = 1
The corresponding values of x are ...
x = 2(-8/5)+1 = -11/5
x = 2(1) +1 = 3
The solutions are (x, y) = (-2.2, -1.6) and (3, 1).
Answer: 6 hours
Step-by-step explanation:
You would first start by making 1 3/5 into a mixed fraction. Getting 8/5. Then you’d divide that by 3/4 and get 6/5. Which then you’d multiply it by 5 and get 6 hours.
Given, the ratio of blocks A, B, C,D are in the ratio 4:7:3:1
Let us consider the common ratio to be ‘x’.
So, toy blocks with alphabet A is 4x and
toy blocks with alphabet B is 7x and
toy blocks with alphabet C is 3x and
toy blocks with alphabet D is x
Again, the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks.
As no. of ‘A’ and ‘C’ blocks are 4x and 3x respectively.
So,
4x=50 + 3x
x=50
Thus, the number of ‘B’ blocks is 7x = 7(50) = 350
350 is the required number.
X=length
y=width
Area (rectangle)=length x width
we suggest this system of equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2 it does not validate
x₂=3/2 ⇒y=x/5=(3/2)/5=3/10
The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles
to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒ 3/10 miles=(3/2) /5 miles
Answer:
2,674.14 g
Step-by-step explanation:
Recall that the formula for radioactive decay is
N = N₀ e^(-λt)
where,
N is the amount left at time t
N₀ is the initial amount when t=0, (given as 42,784 g)
λ = coefficient of radioactive decay
= 0.693 ÷ Half Life
= 0.693 ÷ 18
= 0.0385
t = time elapsed (given as 72 years)
e = exponential constant ( approx 2.7183)
If we substitute these into our equation:
N = N₀ e^(-λt)
= (42,787) (2.7183)^[(-0.0385)(72)]
= (42,787) (2.7183)^(-2.7726)
= (42,787) (0.0625)
= 2,674.14 g
