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3 years ago

How can I solve 2t+4w=6 + 4t +6w= 15 using elimination?

Mathematics

1 answer:

xeze [42]3 years ago

3 0

I hope this helps you

-2/2t+4w=6

-4t-8w= -12

4t+6w=15

-2w=3

w= -3/2

2t-4.3/2=6

2t=12

t=6

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Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year. Whe

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You can see how this works by thinking through what's going on.

In the first year the population declines by 3%. So the population at the end of the first year is the starting population (1200) minus the decline: 1200 minus 3% of 1200. 3% of 1200 is the same as .03 * 1200. So the population at the end of the first year is 1200 - .03 * 1200. That can be written as 1200 * (1 - .03), or 1200 * 0.97

What about the second year? The population starts at 1200 * 0.97. It declines by 3% again. But 3% of what??? The decline is based on the population at the beginning of the year, NOT based no the original population. So the decline in the second year is 0.03 * (1200 * 0.97). And just as in the first year, the population at the end of the second year is the population at the beginning of the second year minus the decline in the second year. So that's 1200 * 0.97 - 0.03 * (1200 * 0.97), which is equal to 1200 * 0.97 (1 - 0.03) = 1200 * 0.97 * 0.97 = 1200 * 0.972.

So there's a pattern. If you worked out the third year, you'd see that the population ends up as 1200 * 0.973, and it would keep going like that.

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3 years ago

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2308127

_______________

Write the expression below in terms of x and y only:

(I'm going to call it "E")

$\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}$

Let

$\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}$

so the expression becomes

$\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}$

• Finding $\mathsf{sin\,\alpha:}$

$\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}$

• Finding $\mathsf{cos\,\alpha:}$

$\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}$

because $\mathsf{cos\,\alpha}$ is positive for $\mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}$

• Finding $\mathsf{cos\,\beta:}$

$\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}$

• Finding $\mathsf{sin\,\beta:}$

$\mathsf{cos^2\,\alpha=y^2}\\\\
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\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}$

because $\mathsf{sin\,\beta}$ is positive for $\mathsf{\beta\in [0,\,\pi].}$

Finally, you get

$\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry

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