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3 years ago

A scatter plot is shown below:

Mathematics

2 answers:

Airida [17]3 years ago

6 0

the 2nd one i think i said i think

ziro4ka [17]3 years ago

5 0

Answer:

(0,0.5) and (10,10) would best make the line.

Step-by-step explanation:

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A die is rolled once what is the probability of rolling not a 4

snow_lady [41]

Answer:

I think 1/6

Step-by-step explanation:

A die has 6 sides so every side has 1/6 chance to land on

8 0

3 years ago

Solutions to -8x + 5 less than or equal to 11

Vilka [71]

Answer:

$x\geq -\frac{3}{4}$

Step-by-step explanation:

$-8x+5\leq 11\\\\1. -8x\leq 11-5\\ -8x\leq 6$

Isolate the parts of the equation with the x variable

$2. \frac{-8}{-8} x\geq\frac{6}{-8} \\x\geq -\frac{3}{4}$

When you divide or multiply by a negative number in inequalities, you must flip the sign of the inequality.

5 0

3 years ago

Read 2 more answers

Which line must be perpendicular to

erma4kov [3.2K]

I think its GI. Could be HJ too. Here's a little reminder about intersecting, parallel, and perpendicular lines...

5 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.

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

Determine if x + 2 is a factor of p(x) = x 4 + 3x 3 + 4x 2 - 8 and explain why.

Andrej [43]

The Factor Theorem says (x - a) is a factor of function p(x) if p(a)=0

so check for p(-2)

= -2^4 +3(-2)^3 + 4(-2)^2 - 8

= 16 - 24 +16 -8

= 0

so (x + 2) is a factor

3 0

3 years ago