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AleksandrR [38]
3 years ago
11

Which ordered pair is a solution to the system of inequalities?

Mathematics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

[see below]

Step-by-step explanation:

I used a graphing program to graph the given system. I also graphed the points given.

Options B's choice lies in the double shaded area, so it would be a solution to the system.

See the graph attached!

Hope this helps you!  

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Aleks [24]
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7 0
3 years ago
100 POINTS
enyata [817]

Answer:

C

Step-by-step explanation:

Given Coordinates:

(-4, -2), (2, -2), (4, 2), (-2, 2)

The shape that uses the coordinates (-4, -2), (2, -2), (4, 2), (-2, 2) would make a parallelogram.

See graph for answer:

So correct choice is C.

8 0
2 years ago
Read 2 more answers
If f(x) = 5x - 7, what is {3)?<br> O A. 22<br> O B. 2<br> O c. 1<br> • D. 8
DerKrebs [107]
<h3>Answer:  8</h3>

Work Shown:

f(x) = 5x - 7

f(3) = 5(3) - 7 .... every x replaced with 3

f(3) = 15 - 7

f(3) = 8

6 0
2 years ago
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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
Solve for x: 5 - (x + 5) &gt; -2(x + 4)
Sav [38]
There you go!!! Hope I helped :)

4 0
2 years ago
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