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ahrayia [7]
3 years ago
15

Can a right triangle be formed using these squares

Mathematics
1 answer:
Ilya [14]3 years ago
7 0

Yes. The two smaller squares have a sum of 169 which is the value of the larger square.

a^2 + b^2 = c^2

25 + 144 = 169

It can be done. Notice the figure below shows you how to arrange the squares to give the answer of a^2 + b^2 = c^2

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A sunken ship is resting at 3,000 feet below sea level. Directly above the ship, a whale is swimming 1,960 feet below sea level.
jeyben [28]

Answer:

   b.

It's too short. Write at least 20 characters to explain it well



5 0
3 years ago
A pill has the shape of a cylinder with a hemisphere at each end. The height of the cylindrical portion is 12mm and the overall
Volgvan

Answer:

The volume of the pill is V=452\ mm^{3}

Step-by-step explanation:

Find the volume of the pill in cubic millimeters

we know that

The volume of the pill is equal to the volume of the cylinder plus the volume of a sphere (two hemisphere is equal to one sphere)

so

we have

V=\frac{4}{3}\pi r^{3}+\pi r^{2}h

r= (18-12)/2=3\ mm\\ h=12\ mm

assume

\pi =3.14

substitute

V=\frac{4}{3}(3.14)(3)^{3}+(3.14)(3)^{2}(12)\\V=452\ mm^{3}

8 0
3 years ago
Billy is playing a game using a bag with 11 marbles inside. The bag contains 3 red, 3 orange, 1 yellow, 2 purple marbles, and 2
Dmitrij [34]

Answer: 3/55

Step-by-step explanation:

From the information given, the bag contains 3 red, 3 orange, 1 yellow, 2 purple marbles, and 2 Pink marbles. Each time he picks an orange marble, she will win a prize.

If he picks a marble the first time, the probability of picking an orange marble will be 3/11. After that we will have 10 marbles left as one has been picked and have 2 orange marbles left, then the probability of picking another orange marble will be 2/10.

Therefore, the probability he will win a prize on both picks will be:

= 3/11 × 2/10

= 6/110

= 3/55

4 0
3 years ago
Will mark brainliest
faust18 [17]

x^3 is strictly increasing on [0, 5], so

\max\{x^3 \mid 0\le x\le5\} = 5^3 = 125

and

\min\{x^3 \mid 0 \le x\le5\} = 0^3 = 0

so the integral is bounded between

\displaystyle \boxed{0} \le \int_0^5x^3\,dx \le \boxed{125}

8 0
2 years ago
Find the area please help
crimeas [40]
8x^2*3xy

24x^3y I believe. Sorry if I’m a little rusty
6 0
3 years ago
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