When comparing the graphs of y = x and y > x, what is a difference in the graphs?

Mathematics

1 answer:

mars1129 [50]2 years ago

8 0

I donnt think that there is a solution, but could be wrong

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Explain why an account balance less than -40 dollars represents a dept greater than 40 dollars

nevsk [136]

You need to pay $40 to make your balance 0 and then another $40 to make it $40.
In all you will have to pay $80

8 0

3 years ago

PLEASE I WILL GIVE BRAINLIEST IF ONLY UR CORRECT.

patriot [66]

Answer:

Mary removed the dollar sign from (X.)

We know this because $4.00x$5.00=$20.00 so that’s correct and we know that $12.50x $4.00=$50.00. So, our only error here is that there is no dollar sign in our (X) numbers. So that is Mary’s mistake. We can correct it by simply adding a dollar sign to 20 to make $20.00 and adding another dollar sign to 50 to make $50.00

Step-by-step explanation:

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2 years ago

For what values of θ is sinθ < cosθ when 0 ≤ θ < π ?

zysi [14]

<h3>Answer: $0 \le \theta < \frac{\pi}{4}$ </h3>

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How to get this answer:

Use the unit circle to note that $\sin\theta = \cos\theta = \frac{\sqrt{2}}{2}$ when $\theta = \frac{\pi}{4}$ (aka 45 degrees)

Beyond this point, cosine is smaller than sine. This means that anything from 0 to pi/4 will have sine be smaller than cosine. It might help to graph y = sin(x) and y = cos(x) on the interval from x = 0 to x = pi.

The two curves y = sin(x) and y = cos(x) intersect at the point $\left(\frac{\pi}{4}, \frac{\sqrt{2}}{2}\right)$

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Here's a more detailed picture of whats going on.

$\sin \theta < \cos \theta\\\\\sin \theta < \sqrt{1-\sin^2\theta}\\\\\sin^2 \theta < 1-\sin^2\theta \\\\2\sin^2 \theta < 1\\\\\sin^2 \theta < \frac{1}{2}\\\\\sin \theta < \sqrt{\frac{1}{2}}\\\\\sin \theta < \frac{1}{\sqrt{2}}\\\\\sin \theta < \frac{\sqrt{2}}{2}\\\\\theta < \arcsin\left(\frac{\sqrt{2}}{2}\right)\\\\\theta < \frac{\pi}{4}\\\\$