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mafiozo [28]
3 years ago
12

If the perimeter of a square is 88 cm, find the length of its diagonal. Give in simplest

Mathematics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

22

Step-by-step explanation:

A square has 4 equal diagonals

And the perimeter of a square is 4L(length)

So the equation will be

=>Perimeter (P) =4L(length)

=>88=4L

=><u>88</u>=<u>4L</u>

4. 4

=>the 4 will divide the 88 to get 22

=>L=22cm

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What is 29 2/19 plus 5 1/2?
boyakko [2]
29 2/19 + 5 1/2 = 29 4/38 + 5 14/38 = 34 18/38 = 34 9/19
7 0
3 years ago
What value of x satisfies this equation? log(5x+20)=2
Ipatiy [6.2K]

Answer:

16

Step-by-step explanation:

Given expression:

       log(5x + 20) = 2

From the given expression, we are to find x;

   Using the rule of logarithm, we can solve this problem;

             logₐb = C

                    b  = Cᵃ

So;

            log(5x + 20) = 2

  this is to base 10;

                    5x + 20 = 10²

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5 0
3 years ago
Read 2 more answers
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.

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Answer:

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Step-by-step explanation:

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