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3 years ago

Solving special systems {y=1/2x+4 {-1/2x+y=-5

Mathematics

1 answer:

ryzh [129]3 years ago

7 0

Answer:

no solution

Step-by-step explanation:

Given the 2 equations

y = $\frac{1}{2}$ x + 4 → (1)

- $\frac{1}{2}$ x + y = - 5 → (2)

Substitute y = $\frac{1}{2}$ x + 4 into (2)

- $\frac{1}{2}$ x + $\frac{1}{2}$ x + 4 = - 5 , that is

4 = - 5 ← not possible

This indicates the system has no solution

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The angle θ 1 is located in Quadrant IV, and cos ⁡ ( θ 1 ) = 9/ 19 , theta, start subscript, 1, end subscript, right parenthesis

Firdavs [7]

Answer:

$sin\theta_1 = - \frac{2\sqrt{70}}{19}$

Step-by-step explanation:

We are given that $\theta_1$ is in <em>fourth</em> quadrant.

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$sin\theta_1$ is always negative in 4th quadrant.

Also, we know the following identity about $sin\theta$ and $cos\theta$ :

$sin^2\theta + cos^2\theta = 1$

Using \theta_1 in place of \theta:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = \frac{9}{19}$

$\Rightarrow sin^2\theta_1 + \dfrac{9^2}{19^2} = 1\\\Rightarrow sin^2\theta_1 = 1 - \dfrac{81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-81}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{280}{361}\\\Rightarrow sin\theta_1 = \sqrt{\dfrac{280}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{2\sqrt{70}}{19}, -\dfrac{2\sqrt{70}}{19}$

$\theta_1$ is in <em>4th quadrant </em>so $sin\theta_1$ is negative.

So, value of $sin\theta_1 = - \frac{2\sqrt{70}}{19}$

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2 years ago

Given h(x) = 5x + 2, solve for a when h(x) = -8.

Whitepunk [10]

Answer:

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Step-by-step explanation:

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11111nata11111 [884]

Answer:

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Step-by-step explanation:

Given that :

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8C4 * 9C5 = (8! /4!4!) * (9! /4!5!)

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Answer:

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