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LuckyWell [14K]

3 years ago

14

Ryan orders a pizza with a 20 inch diameter and asks for a cheese stuff crust along its border . The cheese filled border measur

es blank inches

1 answer:

Kitty [74]3 years ago

5 0

C = pi x d
So the border is about 3.1 x 20 = 62 inches

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Is the ordered pair (4,-3) a solution for the following system of linear inequalities?

Sergio039 [100]

Answer:

no

Step-by-step explanation:

(4, -3) when plugged into the inequality x + y < 1 is not true

4 + (-3) < 1 is not true because 1 is not less than 1

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=3%28%20%7Bx%7D%5E%7B2%7D%20%20%2B%7B3x%7D%5E%7B2%7D%20%20-%202%29" id="TexFormula1" title="3(

Juliette [100K]

The answer should be 12x^2-6

7 0

3 years ago

Read 2 more answers

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.

a)

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:

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

b)

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

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

c)

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.

d)

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

e)

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 me please

MissTica

Answer:

not all of them are proportional because they do not all equal the same.

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Is <img src="https://tex.z-dn.net/?f=1-2x%5E%7B2%7D" id="TexFormula1" title="1-2x^{2}" alt="1-2x^{2}" align="absmiddle" class="l

Schach [20]

Answer:

standard form for a quadratic expression

Step-by-step explanation:

This is a quadratic expression. If you want this expression in standard form, write the terms in descending order of powers of x: -2x^2 + 1.

This -2x^2 + 1 has not yet been factored.

-2x^2 + 1 is in standard form.

7 0

2 years ago