What is the value of x in the solution to the system x+y=10 x-y=4

Mathematics

1 answer:

ANEK [815]2 years ago

6 0

Answer:

x = 7; y = 3

Step-by-step explanation:

I used the Trial and Error Method and found that x = 7 and y = 3

=> 7 + 3 = 10

=> 10 = 10. ✔

=> 7 - 3 = 4

=> 4 = 4. ✔

Therefore, x = 7 and y = 3

Hoped this helped.

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Given: cosθ= -4/5 , sin x = -12/13 , θ is in the third quadrant, and x is in the fourth quadrant; evaluate tan 1/2 θ

mrs_skeptik [129]

Answer:

Step-by-step explanation:

If you're looking for what the half angle of the tangent of theta is, I'm a bit confused as to why you think the angle in the 4th quadrant, x, is relevant. But maybe you don't know it isn't and it's a "trick" to throw you off. Hmm...

Anyways, the half angle identity for tangent is

$tan(\frac{\theta}{2})=\frac{sin\theta}{1+cos\theta}$

There are actually 3 identities for the tangent of a half angle, but this one works just as well as either of the others do, so I'm going with this one.

If theta is in QIII, the value of -4 goes along the x axis and the hypotenuse is 5. That makes the missing side, by Pythagorean's Theorem, -3. Filling in our formula:

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{1+(-\frac{4}{5}) }$ which simplifies a bit to

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{5}{5} -\frac{4}{5} }$ and a bit more to

$tan(\frac{\theta}{2})=\frac{-\frac{3}{5} }{\frac{1}{5} }$

Bring up the lower fraction and flip it to divide to get

$tan(\frac{\theta}{2})=-\frac{3}{5}*\frac{5}{1}$ which of course simplifies to

-3. Choice A.

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3 years ago

Jake rolls a six-sided die, numbered 1 to 6, five times. He rolls a 3 on two attempts.

matrenka [14]

I do believe that the answer to your question is 60%

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3 years ago

If θ=0rad at t=0s, what is the blade's angular position at t=20s

babunello [35]

The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
              ω = 250 (rpm) = 250 * (2π/60) = (25/3)π    rad/s
              ω = 100 (rpm) = 100 * (2π/60) = (10/3)π    rad/s

and from the figure it is clear that the operation is at constant speed but with variable levels
            ⇒   ω = dθ/dt   ⇒   dθ = ω dt

            ∴    θ = ∫₀²⁰ ω dt

while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
       ω = 0                                          from t = 0 to t = 5
       ω = 250 (rpm) = (25/3)π            from t = 5   to t = 15
       ω = 100 (rpm) = (10/3)π            from t = 15 to t = 20

∴ θ = ∫₀⁵ (0) dt   + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt

the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π

∴ θ = 0 + (250/3)π + (50/3)π = 100 π

while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad

Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50