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3 years ago

What is the common factor for 90 and 96?

Mathematics

2 answers:

luda_lava [24]3 years ago

8 0

Answer:

The greatest common factor for 90 and 96 is <u>6</u>.

Step-by-step explanation:

Ierofanga [76]3 years ago

6 0

Answer:

Step-by-step explanation:

Greatest common factor of 90 and 96 is 6.

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what abiotic factor is changing the ecosystem of Lake Washington,and how has this factor changed over time.

raketka [301]

Answer:

Step-by-step explanation:

Temperature is the abiotic factor that is changing the ecosystem. Both the top layer temperature and the overall temperature of the lake are increasing. The warmer temperatures have delayed the yearly springtime mixing of warm and cold water in the lake.

3 0

3 years ago

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

If the ratio of purple bikes to red bikes is 8 to 12 and there are a total of 100 bikes, how many of them are purple? * 40 purpl

Marrrta [24]

Answer:

Step-by-step explanation:

sum the parts of the ratio, 8 + 12 = 20 parts

Divide the total by 20 to find the value of one part of the ratio

100 ÷ 20 = 5 ← value of 1 part of the ratio , thus

8 parts = 8 × 5 = 40 ← number of purple bikes

4 0

3 years ago

Define total compensation

sladkih [1.3K]

In math it means to make a problem easier to work with

6 0

3 years ago

Y >x2-8x+17<br> Please help me !!!:((

Shalnov [3]

I hope this is the correct way to solve one of these.

first you need to know how the graph is so you consider y=x²-8x+17 (you might want to see the picture below to understand in a better way)

then you have to evaluate two points. one above the curve and one underneath the curve to see where y is greater than the curve. so let’s try for instance (0,0) and (4,2)
finally you must determine which points confirm that the statement is actually true. then you select the region for your solution :)

4 0

3 years ago