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3 years ago

Please help, I'm so insanely stupid. ._.

Mathematics

1 answer:

Annette [7]3 years ago

3 0

A) Your primary concerns are the points B and E, so y> .5x+4 and y>or= x-4B) choose one or both points, and enter them into the equations. If the statements are true, then the equations work
for problem C So, any point in the shaded area, but not on the line, are valid points for Natalie's school

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Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago

Please help asap really important

kirill115 [55]

1) slope-2/3, y-intercept-(0,6)
2) slope- -2/3, y-intercept-(0,-3)
3) slope- 8, y-intercept-(0,-6)
4) slope- 6, y-intercept-(0,-8)

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3 years ago

Sharon wants to buy a shirt that costs $60. The sales tax is 5%. How much is the sales tax? What is her total cost for the shirt

Ivahew [28]

Answer:

$3 sales tax

$63 total cost

Step-by-step explanation:

5% of 60 is 3, so that'd be 60 + 3 which is 63, or $63

hope this helps!!!

stay safe!!!

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Show 46/200 on the hundredths grid. then write the fractions as a demical

SSSSS [86.1K]

If we want to has 100 in the denominator we have to divide numerator and denominator by 2. Then total value of fraction doesnt change.
$\frac{46}{200}= \frac{46:2}{200:2}= \frac{23}{100}$
Now read the fraction, its 23 hundreds, now just write as decimal, so, 0.23

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Cheryl has $56 and wants to buy as many notebooks as she can to donate to her school. If each notebook costs $1.60, which inequa

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