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BARSIC [14]
3 years ago
12

Identify the GCF of 10x4y3 − 5x3y2 + 20x2y.

Mathematics
2 answers:
viktelen [127]3 years ago
8 0
<h2>Answer:</h2>

The GCF of the given expression is:

                          5x^2y

<h2>Step-by-step explanation:</h2>

The GCF ( greatest common factor ) is the greatest divisor which divides all the terms completely.

Here we have the expression as:

10x^4y^3-5x^3y^2+20x^2y

Here we see that the factors of the first term are:

10x^4y^3=2\times 5\times x\times x\times x\times x\times y\times y\times y

The factor of the second term is:

-5x^3y^2=-1\times 5\times x\times x\times x\times y\times y

and the factor of the third term is:

20x^2y=2\times 2\times 5\times x\times x\times y

i.e. the factor which is common to all the three terms are:

5\times x\times x\times y=5x^2y

Hence, the GCF is:   5x^2y

Agata [3.3K]3 years ago
7 0
The GCF, or greatest common factor, is the greatest positive integer that evenly multiplies to make the numbers in the set. For simple numbers, GCF(12, 80, 44) is 4.

The strategy for expressions with variables is still to factor out the greatest term.

For 10x^4y^3-5x^3y^2+20x^2y, we look for the largest factor in each term.

For 10x^4y^3, the largest factors are x^4 and y^3. We'll come back to the coefficients in a minute.

For -5x^3y^2, the largest factors are x^3 and y^2. Now, looking back to the previous term and also considering the coefficients, the largest factor of the two is 5x^3y^2. You could rewrite the first term as (5x^3y^2)(2xy) and the second term as -1(5x^3y^2).

Now, we consider the last term, 20x^2y. The largest factor common to the two other terms is 5x^2y.

And that's our final answer for the GCF (answer D).
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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
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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)

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