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2 years ago

Please help I literally don't understand-

Mathematics

2 answers:

Ivenika [448]2 years ago

4 0

Well, QR is the hypotenuse of the triangle and PT is another side of the triangle. You can use phythagoras theorem to work this out. So squareroot 13-8. That should give your your answer

Korolek [52]2 years ago

4 0

You could find QT by using the Pythagorean Theorem.

8^2+QT^2=13^2

64+QT^2=169

QT^2=105

QT=

$\sqrt{105}$

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Triangle TRS is similar to triangle TMN. Angle T = 40°, angle R = 60°, and angle S = 80° . What is the measure of angle M

Andre45 [30]

Answer:

60degrees

Step-by-step explanation:

For similar triangles, the angles of their angles are equal to matter the size.

Hence if triangle TRS is similar to triangle TMN, then;

<R = <M and <S = <N

Given that;

<T = 40 degrees

<R = 60degrees

<S = 80 degrees

Since <R = <M, then <M = 60degrees

7 0

2 years ago

10.125 as percent help please

devlian [24]

10.125 as a percentage is 1012.5%

7 0

3 years ago

Read 2 more answers

160% of what number 56?

Bogdan [553]

Answer:

160% of 35 is 56.

Step-by-step explanation:

$\frac{56}{y} :\frac{160}{100}$

y · 160 = 56 · 100

160y = 5600

160y ÷ 160 = 5600 ÷ 160

y = 35

7 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

The speed of a boat in still water is 40 mph. If it takes the same time for the boat to travel 8 miles upstream as it does to tr

antoniya [11.8K]

Answer:

Current speed is 40mph

Step-by-step explanation:

We firstly need to know what is intended by upstream and downstream. When the boat travels in the direction of the current (horizontal movement of the water), we say it is moving downstream otherwise it is moving upstream.

Let's assume the speed of the boat is A and that of the current is B, then the speed when it moves dowstream would be A + B and the speed if it moves upstream would be A – B.

In this question, we know that the speed of the boat is 40mph. Given that it takes the same time to travel 8miles upstream as it does to travel 24miles downstream, we have

Speed upstream = 8/t

Speed downstream = 24/t

Respectively.

Let speed of current be X, then

Speed downstream = 40 + X

Speed upstream = 40 – X

From the question we have:

(40 + x)/24 = (40 – X)/8

(40 + x)/3 = (40 – x)/1

3(40 – x) = 40 + X

120 – 3x = 40 – X

120 – 40 = 3x – X

2x = 80

X = 40.

Therefore, current speed is 40mph

5 0

3 years ago