Answer:
y = (1/4)x + 0
or
y = (1/4)x
the (1/4) is supposed to be a fraction of 1 in the numerator and 4 in the denominator
Step-by-step explanation:
-- It takes 10 workers 3 days to build 1 house.
so
-- It takes 10 workers 1 day to build 1/3 of a house.
so
-- It takes 1 worker 1 day to build 1/30 of a house.
You got 15 workers and you need 4 houses ?
Well, as we just calculated . . .
-- It takes 1 worker 1 day to build 1/30 of a house.
so
-- It takes 15 workers 1 day to build 15/30 = 1/2 of a house.
so
-- It takes 15 workers 2 days to build 1 whole house.
so
-- It takes 15 workers 8 days to build 4 houses.
Answer:
No Solution
Step-by-step explanation:
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Use the chain rule:
<em>y</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)
<em>y'</em> = sec²(<em>x</em> ² - 5<em>x</em> + 6) × (<em>x</em> ² - 5<em>x</em> + 6)'
<em>y'</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
Perhaps more explicitly: let <em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6, so that
<em>y(x)</em> = tan(<em>x</em> ² - 5<em>x</em> + 6) → <em>y(u(x))</em> = tan(<em>u(x)</em> )
By the chain rule,
<em>y'(x)</em> = <em>y'(u(x))</em> × <em>u'(x)</em>
and we have
<em>y(u)</em> = tan(<em>u</em>) → <em>y'(u)</em> = sec²(<em>u</em>)
<em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6 → <em>u'(x)</em> = 2<em>x</em> - 5
Then
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>u</em>)
or
<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)
as we found earlier.
