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Olenka [21]
3 years ago
10

Solve to find X !!! I have 6 minutes help

Mathematics
2 answers:
ValentinkaMS [17]3 years ago
7 0
I think it’s 22 your answer
Levart [38]3 years ago
5 0
Answer:
22
Explanation:
I-d-k a educated-guess
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Unit rates... please help
harina [27]
When rates are expressed as a quantity of 1 such as 2 feet per second or 5 miles per hour, that is the definition of unit rates.





(Hope this Helps:)
3 0
3 years ago
Write a rule for each sequence. Find the next three terms.<br><br> 8,14,20,26
anastassius [24]
Aruthmetic sequene is
an=a1+(n-1)d
where d=common difference between terms
adds 6 every time
d=6
first term is 8
a1=8
8+6(n-1)
distribute
8+6n-6
8-6+6n
2+6n is answer
8 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.

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
How do you solve -5m+9n= 15, for m
Fed [463]

Answer:

m= 9/5 n - 3

Step-by-step explanation:

Let's solve for m.

−5m+9n=15

Step 1: Add -9n to both sides.

−5m+9n+−9n=15+−9n

−5m=−9n+15

Step 2: Divide both sides by -5.

-5m/-5 = -9n+15/-5

m= 9/5 n - 3

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D%20%284%20-%203x%29%20%3D%20%5Cfrac%7B1%7D%7B7%7D%20%283x%20-%204%29" id=
HACTEHA [7]

Answer:

1×(4-3x)/5=1×(3x×4)/7

(4-3x)/5=(3x-4)/7

7(4-3x)=5(3x-4)

28-21x=15x+21x

28+20=15x+21x

48=36x

48/36=x

x=4/3

6 0
3 years ago
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