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3 years ago

Sin θ > 0 and tan θ < 0 then:

Mathematics

1 answer:

andriy [413]3 years ago

4 0

Answer:

last choice

Step-by-step explanation:

sin(theta) is positive

tan(theta)=sin(theta)/cos(theta) is negative

If sin(theta) is positive and tan(theta) is negative, then cos(theta) is negative because +/-=-

So we are looking for the quadrant where sine is positive and cosine is negative or when x is negative and y is positive.

This is in quadrant 2 so our angle theta is between 90 and 180 degrees.

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An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

natima [27]

Answer:

Speed of the plane in still air: $779\; {\rm km \cdot h^{-1}}$ .

Windspeed: $100\; {\rm km \cdot h^{-1}}$ .

Step-by-step explanation:

Assume that $x\; {\rm km \cdot h^{-1}}$ is the speed of the plane in still air, and that $y\; {\rm km \cdot h^{-1}}$ is the speed of the wind.

When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be $(x - y)\; {\rm km \cdot h^{-1}}$ .
When this plane is travelling in the same direction as the wind, the ground speed of this plane would be $(x + y)\; {\rm km \cdot h^{-1}}$ .

The question states that when going against the wind ( $v = (x - y)\; {\rm km \cdot h^{-1}}$ ,) the plane travels $6111\; {\rm km}$ in $9\; {\rm h}$ . Hence, $9\, (x - y) = 6111$ .

Similarly, since the plane travels $7911\; {\rm km}$ in $9\; {\rm h}$ when travelling in the same direction as the wind ( $v = (x + y)\; {\rm km \cdot h^{-1}}$ ,) $9\, (x + y) = 7911$ .

Add the two equations to eliminate $y$ . Subtract the second equation from the first to eliminate $x$ . Solve this system of equations for $x$ and $y$ : $x = 779$ and $y = 100$ .

Hence, the speed of this plane in still air would be $779\; {\rm km \cdot h^{-1}}$ , whereas the speed of the wind would be $100\; {\rm km \cdot h^{-1}}$ .

3 0

2 years ago

Given this function: f ( x )=−2cos(4 πx ) Find the following:

Neporo4naja [7]

Answer:

Period =½

Equation of midline, y=0

Maximum =2

Minimum=-2

Step-by-step explanation:

The given function is

$f(x) = - 2 \cos(4 \pi \: x)$

The period is given by:

$\frac{2\pi}{b} = \frac{2\pi}{4\pi} = \frac{1}{2}$

The equation of the midline is y=0 since there is no vertical shift

The amplitude of this function is 2 so the range is -2≤y≤2.

Hence the maximum value is 2 and minimum value is -2

4 0

4 years ago

Which formula gives the zeros of y = sin(x)

mojhsa [17]

Trigonometric formula

7 0

4 years ago

Making use of the fact hat eq. (6.20) is an exact differential expression, show that what is the result of application of this e

nevsk [136]

Since there is no figure attached, I will describe the derivation of the ideal gas law. The combined gas law has no official founder; it is simply the incorporation of the three laws that was discovered. The combined gas law is a gas law that combines Gay-Lussac’s Law, Boyle’s Law and Charle’s Law. Boyle’s law states that pressure is inversely proportional with volume at constant temperature. Charle’s law states that volume is directly proportional with temperature at constant pressure. And Gay-Lussac’s law shows that pressure is directly proportional with temperature at constant volume. The combination of these laws known now as combined gas law gives the ratio between the product of pressure-volume and the temperature of the system is constant. Which gives PV/T=k(constant). When comparing a substance under different conditions, the combined gas law becomes P1V1/T1 = P2V2/T2.

6 0

4 years ago

Find an equation of the tangent line to the hyperbola: x^2/a^2 - y^2/b^2 = 1 at the point (x0,x1)

musickatia [10]

The standard form of a hyperbola is x2/a2 − y2/ b2 = 1
the tangent line is the first derivative of the functiony′ = b^2x/ a^2 y
hence the slope is m = b^2 x0 / a^2 x1

Therefore the equation of the tangent line isy−x1 = b^2 x0 / a^2 x1* (x−x0)

8 0

3 years ago