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Anna [14]
2 years ago
11

Will mark brainlist if right

Mathematics
1 answer:
dlinn [17]2 years ago
7 0

Answer: C.

Step-by-step explanation:

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If y=-4+11 and 3x+y=9 what is the value of y
Kruka [31]

y=-4+11=7\\y=7\\3x+7=9\\x=2/3

7 0
3 years ago
Megan has been planting young trees in her garden. One maple tree is 66 centimeters tall and is growing 8 centimeters per month.
EastWind [94]
Once per month there is food that is 33 centimeters long
7 0
2 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
What is the midpoint of the segment shown below?
kolbaska11 [484]

The midpoint of the segment is (-15/2, -15/2)

<h3>How to determine the midpoint?</h3>

The complete question is in the attached image

The points are given as:

(-8, -7) and (-7, -8)

The midpoint is calculated as:

(x,y) = 1/2 * (x1 + x2, y1 + y2)

So, we have:

(x,y) = 1/2 * (-8 - 7, -7 - 8)

Evaluate the difference

(x,y) = 1/2 * (-15, -15)

Evaluate the product

(x,y) = (-15/2, -15/2)

Hence, the midpoint of the segment is (-15/2, -15/2)

Read more about midpoints at:

brainly.com/question/4747771

#SPJ1

8 0
2 years ago
Solve this n get pts :)​
Simora [160]
D+100+40=180
d=180-100-40=40 (sum of all angles)
e= 100+40= 140 (exterior angle)
180-140=40
C+50+40=180
C=180-50-40= 90 (right angle)
A=50
B=180-90-50= 40
8 0
2 years ago
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