Answer:
x=67.4
Step-by-step explanation:
well sin30:1/2
So y is not 30 since 5/13 doesn’t equal 1/2
So it’s either the the third or fourth option
And x>y
So option 4 is correct
Answer:
y = (3/5)x+26/5 or 5y = 3x+26
Step-by-step explanation:
Applying,
The equation of a line in two point form
(y₂-y₁)/(x₂-x₁) = (y-y₁)/(x-x₁)............... Equation 1
From the question,
Given: y₂ = 7, y₁ = 4, x₂ = 3, x₁ = -2
Substitute these values into equation 1
(7-4)/[3-(-2)] = (y-4)/(x+2)
3/5 = (y-4)/(x+2)
5(y-4) = 3(x+2)
5y-20 = 3x+6
5y = 3x+6+20
5y = 3x+26
y = (3/5)x+26/5
Hence the equation of the line is y = (3/5)x+26/5 or 5y = 3x+26
Answer:
Steve
Step-by-step explanation:
steve is buying 200 minutes which is more than 150 minutes
Hope i helped :) :) :)
If
is the amount of strontium-90 present in the area in year
, and it decays at a rate of 2.5% per year, then

Let
be the starting amount immediately after the nuclear reactor explodes. Then

or simply

So that after 50 years, the amount of strontium-90 that remains is approximately

or about 28% of the original amount.
We can confirm this another way; recall the exponential decay formula,

where
is measured in years. We're told that 2.5% of the starting amount
decays after 1 year, so that

Then after 50 years, we have
