Factor the algebraic expression. 24a + 15 24a + 15 =
2 answers:
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Answer:
Allison worked 6 hours lifeguarding and 3 hours washing cars.
Step-by-step explanation:
Let
Number of hours Allison worked lifeguarding last week = x
Number of hours Allison worked washing cars last week = y
1. Last week Allison worked 3 more hours lifeguarding than hours washing cars hours, then
<u>Lifeguarding:</u>
$12 per hour
$12x in x hours.
<u>Washing cars:</u>
$8 pere hour
$8y in y hours.
2. Allison earned a total of $96, hence
You get the system of two equations:
Plot the graphs of these two equations (see attached diagram). These line intersect at point (6,3), so Allison worked 6 hours lifeguarding and 3 hours washing cars.
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)
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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)
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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.
There are 138 female patients in the hospital.
<h3>Solving word problems involving ratios.</h3>The solution to the word problem involving ratios in the given problem can be determined as follows.
From the given information:
The total number of people in the hospital = male + female
= 184 + x(unknown)
- If the ratio of male to female is 4: 3.
- The total ratio is 4+3 = 7
The ratio of males in the hospital is:
x = 322
The number of female patients in the hospital can now be computed as:
= 138 female patients
Check:
The total number of people in the hospital is:
= 184 + 138
= 322 patients
Learn more about word problems here:
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