What is the value of x in the equation 2x+3y=36 when y=6
1 answer:
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Answer:
$41269.983
rounded:
$41,269.98
Step-by-step explanation:
well you begin with the equation a=p(1+r/n)
so 30,000(1+0.08/12)to the power of 4 times 12
when you plug all of that in a calculator your receive $41269.983
All the answers to the given parts are mentioned below -
<h3>What is the general equation of a Straight line? How it represents a proportional relationship?</h3>The general equation of a straight line is -
[y] = [m]x + [c]
where -
[m] is slope of line which tells the unit rate of change of [y] with respect to [x].
[c] is the y - intercept i.e. the point where the graph cuts the [y] axis.
y = mx also represents direct proportionality. We can write [m] as -
m = y/x
OR
y₁/x₁ = y₂/x₂
We have a cafeteria serves lemonade that is made from a powdered drink mix. There is a proportional relationship between the number of scoops of powdered drink mix and the amount of water needed to make it. For every 2 scoops of mix, one-half gallon of water is needed, and for every 6 scoops of mix, one and one-half gallons of water are needed.
We can write the proportional relationship as -
y = kx
Now, from the given information, we can write -
2 scoops need 0.5 gallon of water
6 scoops need 1.5 gallon of water
So -
k = 2/0.5
k = 2/(1/2)
k = 2 x 2 = 4
Equations that represents the relationship can be written as -
y = 4x + c
Now, 2 scoops need 0.5 gallon of water.
2 = 4 x 1/2 + c
2 = 2 + c
c = 0
So, the equation will be y = 4x.
Graph of y = 4x is attached at the end.
For 10 scoops of water -
10 = 4x
x = 2.5 gallons of water is needed.
Therefore, all the answers to the given parts are mentioned above.
To solve more questions on proportional relationship, visit the link below-
#SPJ1
Answer:
a) d(sinh(f(x)))/dx = cosh(f(x))·df(x)/dx
b) d(cosh(f(x))/dx = sinh(f(x))·df(x)/dx
c) d(tanh(f(x))/dx = sech(f(x))²·df(x)/dx
d) d(sech(4x+2))/dx = -4sech(4x+2)tanh(4x+2)
Step-by-step explanation:
To do these, you need to be familiar with the derivatives of hyperbolic functions and with the chain rule.
The chain rule tells you that ...
(f(g(x)))' = f'(g(x))g'(x) . . . . where the prime indicates the derivative
The attached table tells you the derivatives of the hyperbolic trig functions, so you can answer the first three easily.
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a) sinh(u)' = sinh'(u)·u' = cosh(u)·u'
For u = f(x), this becomes ...
sinh(f(x))' = cosh(f(x))·f'(x)
__
b) After the same pattern as in (a), ...
cosh(f(x))' = sinh(f(x))·f'(x)
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c) Similarly, ...
tanh(f(x))' = sech(f(x))²·f'(x)
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d) For this one, we need the derivative of sech(x) = 1/cosh(x). The power rule applies, so we have ...
sech(x)' = (cosh(x)^-1)' = -1/cosh(x)²·cosh'(x) = -sinh(x)/cosh(x)²
sech(x)' = -sech(x)·tanh(x) . . . . . basic formula
Now, we will use this as above.
sech(4x+2)' = -sech(4x+2)·tanh(4x+2)·(4x+2)'
sech(4x+2)' = -4·sech(4x+2)·tanh(4x+2)
_____
Here we have used the "prime" notation rather than d( )/dx to indicate the derivative with respect to x. You need to use the notation expected by your grader.
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<em>Additional comment on notation</em>
Some places we have used fun(x)' and others we have used fun'(x). These are essentially interchangeable when the argument is x. When the argument is some function of x, we mean fun(u)' to be the derivative of the function after it has been evaluated with u as an argument. We mean fun'(u) to be the derivative of the function, which is then evaluated with u as an argument. This distinction makes it possible to write the chain rule as ...
f(u)' = f'(u)u'
without getting involved in infinite recursion.
