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PSYCHO15rus [73]
3 years ago
13

If csc θ = -5/4 on the interval (270°, 360°), find tan θ.

Mathematics
1 answer:
adell [148]3 years ago
6 0
I believe the answer is 3
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If a rectangle has an area of 36 square units, what could the dimensions be?
Oxana [17]

The area of a rectangle is length times width. Since a square's length is equal to its width, a square's area is equal to the length of one side times itself. So, reversed, the square root of the square's area gives the length of one side. In this case, the square root of 36 square centimeters is 6 centimeters. Again, since all four sides of a square are the same, its perimeter is equal to the length of one side times four. For a square with one side equal to 6 centimeters, the perimeter equals 24 centimeters.

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3 years ago
Help ill give brainlest pls
Mamont248 [21]

Answer:

The answer is C. The translation of a line is a pair of parallel lines.

Step-by-step explanation:

A, B, and D are true. C is <em>not</em> true, and therefore the correct answer.

5 0
2 years ago
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Mily recycled 66 cans in February, 94 cans in March, 122 cans in April, and 150 cans in May. If this pattern went from January t
Mice21 [21]

Answer:

38 cans

Step-by-step explanation:

while looking at the numbers given I noticed that the number of cans recycled went up by 28 cans, so I took February's number, 66, and subtracted 28 to get 38

6 0
3 years ago
Please help with my homework
matrenka [14]

It's 65. Because you add the other angles and get 90+25 and get 115. 180 -115 = 65.

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3 years ago
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Consider the region, R, bounded above by f(x)=x2−6x+9 and g(x)=−3x+27 and bounded below by the x-axis over the interval [3,9]. F
Salsk061 [2.6K]

Answer:

22.5

Step-by-step explanation:

The region R contains every point of the plane with coordinate x between 3 and 9, and with coordinate y positive such that y < f(x) and y < g(x).

We can note that both f and g are positive on [3,9] because g is a decreasing linear function and g(9) = 0, thus g is positive in every other point of the interval, and f(x) = (x-3)^2 is always positive excpept when x = 3, where it reaches the value 0.

The interception of the graphs takes place for a value x such that f(x) = g(x).

We compute x^2-6x+9 = -3x + 27, thus x^2-6x+9-(-3x + 27) = x^2-3x -18 = 0.

The roots of that quadratic function are

r_1, r_2 = \frac{3 ^+_- \sqrt { 9 +72}}{2} = \frac{3^+_-9}{2} , thus r1 = 6, r2 = -3. We dont care about -3 because it is outside the interval, but we know that f and g graphs intersects on x = 6. Thus, we obtain, due to Bolzano Theorem:

  • On the interval [3,6), the function f in smaller because it takes the value 0 on x=3, while g is always positive.
  • On the interval (6,9]. the function g is smaller because it takes the value 0 on x=9, while f is always positive

Hence, the upper bound is f on the interval [3,6) and g on the interval (6,9]. While the lower bound is the 0 function.

We need to calculate the following integral, using Barrow's rule

\int\limits^6_3 {x^2-6x+9} \, dx + \int\limits^9_6 {-3x+27} \, dx = (\frac{x^3}{3} - 3x^2 + 9x) |^6_3 + (\frac{-3x^2}{2} + 27x)|^9_6 = \\  (18 - 9) + (121.5-108) = 22.5

As a result, the area of the region R is 22.5

6 0
3 years ago
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