All categories

2 years ago

Find out the sum by suitable arrangement. 1 + 2 + 3 + 4 + 5 + 95 + 96 + 97 + 98 + 99

Mathematics

2 answers:

Ksivusya [100]2 years ago

7 0

1 + 2 + 3 + 4 + 5 + 95 + 96 + 97 + 98 + 99. = 500

hoa [83]2 years ago

4 0

Answer:

500

Step-by-step explanation:

rearrange the sum as

(99 + 1) + (98 + 2) + (97 + 3) + (96 + 4) + (95 + 5)

= 100 + 100 + 100 + 1000 + 100

= 5 × 100

= 500

You might be interested in

What number would complete the pattern below<br> 16 4 12 36 9 27 44 11

Liono4ka [1.6K]

33 is the answer :) Please give me the brainliest answer, and a rate and thanks.

5 0

3 years ago

Use cosine to find the missing value round to 2 decimal places #trigonometry

Naddika [18.5K]

Answer:

1. 17.27 cm

2. 19.32 cm

3. 24.07°

4. 36.87°

Step-by-step explanation:

1. Determination of the value of x.

Angle θ = 46°

Adjacent = 12 cm

Hypothenus = x

Using cosine ratio, the value of x can be obtained as follow:

Cos θ = Adjacent /Hypothenus

Cos 46 = 12/x

Cross multiply

x × Cos 46 = 12

Divide both side by Cos 46

x = 12/Cos 46

x = 17.27 cm

2. Determination of the value of x.

Angle θ = 42°

Adjacent = x

Hypothenus = 26 cm

Using cosine ratio, the value of x can be obtained as follow:

Cos θ = Adjacent /Hypothenus

Cos 42 = x/26

Cross multiply

x = 26 × Cos 42

x = 19.32 cm

3. Determination of angle θ

Adjacent = 21 cm

Hypothenus = 23 cm

Angle θ =?

Using cosine ratio, the value of θ can be obtained as follow:

Cos θ = Adjacent /Hypothenus

Cos θ = 21/23

Take the inverse of Cos

θ = Cos¯¹(21/23)

θ = 24.07°

4. Determination of angle θ

Adjacent = 12 cm

Hypothenus = 15cm

Angle θ =?

Using cosine ratio, the value of θ can be obtained as follow:

Cos θ = Adjacent /Hypothenus

Cos θ = 12/15

Take the inverse of Cos

θ = Cos¯¹(12/15)

θ = 36.87°

8 0

3 years ago

Explain truth tables and how they work

lions [1.4K]

A truth table is a way of organizing information to list out all possible scenarios. We title the first column p for proposition. In the second column we apply the operator to p, in this case it's ~p (read: not p). So as you can see if our premise begins as True and we negate it, we obtain False, and vice versa.

6 0

3 years ago

Read 2 more answers

A drawer contains 12 brown socks and 12 black socks, all unmatched. A man takes socks out at random in the dark. 21. Required in

avanturin [10]

Answer:

6 socks

Step-by-step explanation:

What we must do is calculate the probability of this happening, that he takes out two black socks in the first two taken out.

There are 12 black socks and in total they are 24, therefore the probability of drawing 1 is:

12/24

and now the probability of getting another one is 11 (there is one less outside) and in total they are 23:

11/23

the final probability is the multiplication of these events:

(12/24) * (11/23)

P = 0.24

Now, to know how many you should get, we multiply the probability by the total number of socks, that is:

0.24 * 24 = 5.76

So you must take out at least 6 socks for the above to happen.

6 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