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Sindrei [870]
2 years ago
10

What is the solution to the equation? 78=m32

Mathematics
1 answer:
tekilochka [14]2 years ago
6 0

Step-by-step explanation:

78 = m32

m = 78/32 = 39/16 = 2.4375

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What is 6x-13<6(x-2)
ohaa [14]

For this case we have the following inequality:

6x-13

To find the solution we follow the steps below:

We apply distributive property on the right side of inequality:

6x-13

Adding 13 to both sides of the inequality we have:

6x

We subtract 6x on both sides of the inequality:

0

Thus, we have that any value of "x" makes the inequality fulfilled. Thus, the solution is given by all real numbers.

Answer:

The solution set is (-∞,∞)

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3 years ago
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Explanation:

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3 years ago
Find the absolute maximum and absolute minimum values of f on the given interval. f(t) = 9t + 9 cot(t/2), [π/4, 7π/4]
agasfer [191]

Answer:

the absolute maximum value is 89.96 and

the absolute minimum value is 23.173

Step-by-step explanation:

Here we have cotangent given by the following relation;

cot \theta =\frac{1 }{tan \theta} so that the expression becomes

f(t) = 9t +9/tan(t/2)

Therefore, to look for the point of local extremum, we differentiate, the expression as follows;

f'(t) = \frac{\mathrm{d} \left (9t +9/tan(t/2)  \right )}{\mathrm{d} t} = \frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2}

Equating to 0 and solving gives

\frac{9\cdot sin^{2}(t)-\left (9\cdot cos^{2}(t)-18\cdot cos(t)+9  \right )}{2\cdot cos^{2}(t)-4\cdot cos(t)+2} = 0

t=\frac{4\pi n_1 +\pi }{2} ; t = \frac{4\pi n_2 -\pi }{2}

Where n_i is an integer hence when n₁ = 0 and n₂ = 1 we have t = π/4 and t = 3π/2 respectively

Or we have by chain rule

f'(t) = 9 -(9/2)csc²(t/2)

Equating to zero gives

9 -(9/2)csc²(t/2) = 0

csc²(t/2)  = 2

csc(t/2) = ±√2

The solutions are, in quadrant 1, t/2 = π/4 such that t = π/2 or

in quadrant 2 we have t/2 = π - π/4 so that t = 3π/2

We then evaluate between the given closed interval to find the absolute maximum and absolute minimum as follows;

f(x) for x = π/4, π/2, 3π/2, 7π/2

f(π/4) = 9·π/4 +9/tan(π/8) = 28.7965

f(π/2) = 9·π/2 +9/tan(π/4) = 23.137

f(3π/2) = 9·3π/2 +9/tan(3·π/4) = 33.412

f(7π/2) = 9·7π/2 +9/tan(7π/4) = 89.96

Therefore the absolute maximum value = 89.96 and

the absolute minimum value = 23.173.

7 0
3 years ago
Please explain how to graph this and how you did it
inysia [295]
So you would start at the origin which is 0 then go down one cause of the -1 and right 2 because of the positive 2
3 0
3 years ago
Solving the following system using the elimination of substitution method.<br> 5a+3b=11<br> -2a+3=b
PSYCHO15rus [73]

Answer:

a = -2

b = 7

Step-by-step explanation:

5a+3b=11

-2a+3=b

substitute for b:

5a + 3(-2a + 3) = 11

5a -6a + 9 = 11

-a = 2

a = -2

b = 7

4 0
2 years ago
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