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DerKrebs [107]
2 years ago
7

5th grade math. correct answer will be marked brainliest

Mathematics
2 answers:
irina1246 [14]2 years ago
7 0

Answer:

sorry but the picture if you put one on here isn't showing up maybe Repost and I could answer

gulaghasi [49]2 years ago
7 0
Where is the question
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Plizz help me out on this oneee
Kobotan [32]
There are 30 students all together, and 4 of them are Freshman girls (Freshwomen ?). So the probability that one randomly chosen student is a Female Frosh is 4/30 = 2/15.
8 0
3 years ago
(9 x 10) + (20x + 21) = 571<br><br> What does (x) equal
creativ13 [48]
(9 × 10) + (20x + 21) = 571
90 + 20x + 21 = 571
111 + 20x = 571
- 111
20x = 460
÷ 20
x = 23
I hope this helps!
8 0
3 years ago
Read 2 more answers
Will mark brainliest please help <br>​
snow_lady [41]

Answer:8_1

Step-by-step explanation:

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.

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
A man sold a fan for Rs. 400. Find the cost price if he incurred a loss of 15%.
statuscvo [17]

Answer:

Cost price of the fan is \text {Rs. 470.59}.

Step-by-step explanation:

Given: A man sold a fan for  \text{Rs.}\:400. He incurred a loss of 15\%.

To find: The cost price of the fan

Solution:

We have,

Selling price, S.P.,  of the fan = \text{Rs.} 400

Loss =15\%

We know that \text {Loss\%}=\frac{\text{C.P.-S.P.}}{\text{C.P.}}\times100\%

Now let the C.P. of the fan be \text{Rs.} \:x

So, 15=(\frac{x-400}{x})\times 100

\implies 15x=100x-40000

\implies 100x-15x=40000

\implies 85x=40000

\implies x=\frac{40000}{85}

\implies x= 470.59

Hence, the cost price of the fan is \text {Rs. 470.59}.

7 0
3 years ago
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