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lbvjy [14]
3 years ago
12

30 - 40

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

10

10

10

10

ZU80

10

Hahaha jokeeeee

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If (tan^3 theta -1) / (tan theta - 1) - sec^2 theta +1 = 0, find cot theta.
stiv31 [10]
Hey there !

Check the attachment.
Hope it helps you :)

7 0
3 years ago
Read 2 more answers
Sam is paid $14 per hour plus 6% of sales. If he worked 37.5 hours in one week, what would his sales need to be for him to make
Ulleksa [173]

Answer:

Step-by-step explanation

Let the sales be represented by x

Amount made from no of hours worked = $14 × 37.5 hours=$525

Sales made = 6% of x=0.06x

Total = 0.06x+$525=$733.50

0.06x=733.50-$525=$208.5

x=$3,475

4 0
3 years ago
Let f be the function defined by f(x) = 3x^5-5x^3+2
sveticcg [70]
(a) When f is increasing the derivative of f is positive.

    f'(x) = 15x^4 - 15x^2 > 0
            15x^2(x^2 - 1)> 0
               x^2 - 1    > 0 (The inequality doesn't flip sign since x^2 is positive)
               x^2 > 1
    Then f is increasing when x < -1 and x > 1.

(b) The f is concave upward when f''(x) > 0.
    f''(x) = 60x^3 - 30x > 0
           30x(2x^2 - 1) > 0
             x(2x^2 - 1) > 0
            x(x^2 - 1/2) > 0
x(x - 1/sqrt(2))(x + 1/sqrt(2)) > 0

    There are four regions here. We will check if f''(x) > 0.
    x < -1/sqrt(2):     f''(-1) = -30 < 0
    -1/sqrt(2) < x < 0: f''(-0.5) = 7.5 > 0
    0 < x < 1/sqrt(2):  f''(0.5) = -7.5 < 0
    x > 1/sqrt(2):      f''(1) = 30 > 0

    Thus, f''(x) > 0 at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).
    Therefore, f is concave upward at -1/sqrt(2) < x < 0 and x > 1/sqrt(2).

(c) The horizontal tangents of f are at the points where f'(x) = 0
    15x^2(x^2 - 1) = 0
    x^2 = 1
    x = -1 or x = 1
    f(-1) = 3(-1)^5 - 5(-1)^3 + 2 = 4
    f(1) = 3(1)^5 - 5(1)^3 + 2 = 0

    Therefore, the tangent lines are y = 4 and y = 0.
8 0
4 years ago
Read 2 more answers
Does this diverge or converge?
olga2289 [7]

Answer:

L = 7/6 , and the correct option is B, the series diverges.

Step-by-step explanation:

Here we have that:

an = \frac{(3n + 3)*7^n}{6^{n+4}}

Now we want to see the quotient test, then we need to do find the value of the quotient as n tends to infinity. Then we first need to find the quotient:

a(n + 1)/a(n) = \frac{(3(n + 1) + 3)*7^{n + 1}}{6^{n + 1 + 4}}*\frac{6^{n + 4}}{(3*(n) + 3)*7^{n}}

We can simplify the right part to get:

\frac{(3(n + 1) + 3)*7^{n + 1}}{6^{n + 1 + 4}}*\frac{6^{n + 4}}{(3*(n) + 3)*7^{n}}=  \frac{(3n + 3 + 3)*7^{n + 1}}{6^{n + 5}}*\frac{6^{n + 4}}{(3*n + 3)*7^{n}}

\frac{(3n + 3 + 3)*7^{n + 1}}{6^{n + 5}}*\frac{6^{n + 4}}{(3*n + 3)*7^{n}}= \frac{(3n + 3 + 3)*7}{(3*n + 3)*6}

Now we want to find the limit of this as n goes to infinity, we can see that both parts will tend to infinity, then we need to use the Lhopital rule and find the limit of the quotient of the first derivatives.

The first derivatives are:

For the denominator: (3*6)

For the numerator: (3*7)

Then we need to find the limit as n goes to infinity to:

\frac{3*7}{3*6} = \frac{7}{6}

But this does not depend on the value of n, then the limit is just 6/7.

L = 7/6

And 7/6 > 1.

Then we can conclude that our series diverges.

3 0
3 years ago
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
Reil [10]
F(x) = 8/x and g(x) = 8/x

To find f(g(x)), all you do is substitute g(x) into the x value of f(x). 

So f(g(x)) = 8/(8/x) which = x

Similarly g(f(x)) is found by substituting f(x) into the x value of g(x)

g(f(x)) = 8/(8/x) which = x
6 0
4 years ago
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