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UkoKoshka [18]
3 years ago
11

a farmer sold 2/5 of his cows and another 10th was stolen he then gave 20% of the cows that was left what percentage of the cows

is left​
Mathematics
1 answer:
igor_vitrenko [27]3 years ago
8 0

Answer:

4/5 of the cattle is left

Step-by-step explanation:

Given,

A farmer sells 2/5 of his cattle

Let "x" be the total number of cattle

Then,

sold = 2 / 5 x

Then,

Remaining = x - 2x / 5

remaining = 3x / 5

He gives 1/3 of the remainder to his son

son = 1 / 3 × 3x / 5

son = x / 5

Then,

Remaining = 3x / 5 - x / 5

Remaining = 4x / 5

Hope this answer helps you :)

Have a great day

Mark brainliest

Thus 4/5 of the cattle is left

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Select the equation in which the graph of the line has a negative slope and the y intercept equals -5
tatuchka [14]

Answer:

Step-by-step explanation:

Next time, please share the possible answer choices.

Since the slope and y-intercept are given here, we can immediately write out the slope-intercept form of the equation of this line:

y = mx + b becomes y = -1x - 5, where I have arbitrarily chosen the negative slope -1.

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3 years ago
What value of x makes the following equation true 4+3x=160
Alexxandr [17]

Answer:

x=52

Step-by-step explanation:

160=4+3x

Subtract 4 from both sides

156=3x

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3 years ago
What is the least common denominator for the fractions 1/3and2/9?
Mice21 [21]

Answer:

LCD= 9

Step-by-step explanation:

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2 years ago
Two families went to the event. The first family bought two children's tickets and one adult ticket and paid a total of 8.2 euro
blagie [28]

Answer:

Step-by-step explanation:

step a system of two equations         c = child ticket     a = adult ticket

eq 1)  2c + 1a = 8.2            multiply by 2

eq 2) 3c + 2a = 14.1

I will multiply eq 1 times TWO and subtract eq 2 from eq 1a)

         eq 1a)  4c + 2a = 16.4

          eq 2)  3c + 2a = 14.1

subtract  (4c - 3c) + (2a -2a)  = 16.4 - 14.1  

                    c      +       0       =   2.3 euros     for one child ticket

Now find the adult ticket price,  plug 2.3 for c into eq 1)

     eq 1)  2c + 1a = 8.2

     eq 1)  2(2.3) + 1a = 8.2         solve for a

                 4.6  + a   =  8.2         substract 4.6 from both sides

                             a =  8.2 - 4.6

                               =   3.6 euros    for one adult ticket

double check using eq 2)    we know c and a values

    eq 2)  3c + 2a = 14.1

    eq 2)  3(2.3) + 2(3.6) = 14.1

                   6.9 + 7.2    =  14.1

                          14.1    =  14.1

5 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
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