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2 years ago

9

2. In a class 70 students, 44 like Bookkeeping, 36 like- Commerce, and 17 like both bookkeeping and commere How many student lik

e both book keeping and commerce?

2 answers:

tigry1 [53]2 years ago

5 0

Answer:

17

Step-by-step explanation:

you said that in the problem lol

Varvara68 [4.7K]2 years ago

5 0

Answer:

17 like both bookkeeping and commerce

Step-by-step explanation:

The problem statement itself tells you ...

"17 like both bookkeeping and commerce"

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Find the distance between (3,4) and (4,-6) if necessary, round to the nearest tenth.

Ratling [72]

Answer:

The distance is:

d = 10.0 units (Rounded to the nearest the Tenths Place)

Step-by-step explanation:

Given the points

(3,4)

(4,-6)

The distance 'd' between (3,4) and (4,-6)

$\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):$

$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

substituting the points values

$=\sqrt{\left(4-3\right)^2+\left(-6-4\right)^2}$

$=\sqrt{1+10^2}$

$=\sqrt{1+100}$

$=\sqrt{101}$

$=10.0$ units (Rounded to the nearest the Tenths Place)

Thus, the distance is:

d = 10.0 units (Rounded to the nearest the Tenths Place)

4 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)

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3 years ago

Write each fraction as a decimal. Then identify each decimal as terminating or repeating.

cluponka [151]

1/6 = 0.16
1/8 = 0.125
1/11 = 0.09
2/9 = 0.22
4/5 = 0.8
5/9 = 0.56
1/2 = 0.5
7/9 = 0.78

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3 years ago

Bethany s making trail mix with 3 cups of raisins for every 2 cups of peanuts

DaniilM [7]

Answer:

4 cups of raisins

3 cups of dates

6 cups of peanuts

and 2 cups of cashews

ratio cashews to raisins= 2:4

= 1:2

ratio dates to peanuts= 3:6

= 1:2

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4 years ago

True Statements Plz help me

Ket [755]

Answer:

Bvcvbcc

Step-by-step explanation:

Bbvvbbbbb

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3 years ago