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olga nikolaevna [1]
3 years ago
15

Please help! Help me with the steps as well not just answers:/

Mathematics
1 answer:
serious [3.7K]3 years ago
3 0
For question 1 you need to find the sqared values of the hypotenuse, which has a length of 15, and leg b, which has a length of 12. This gives you a hypotenuse of 225 and a leg which is 144. Subtract your given leg squared from your hypotenuse squared (aka 225-144). This leaves you with 81. Now find the square root of this, which is 9.

For question 2 you need to find the squared value of legs a and b. Leg a has an initial length of 6, which squared is 36. The same applies for leg b. Now add 36 to 36 to get 72. All you have to do is find the square root of that to have your answer. Since 72 isn't a perfect square, just simplify the root to 3 root 8, aka
3 \sqrt{8}
For question 3, repeat the process used in question 1. The square of a square root is the original number (ex: the sqare root of 3 squared is 3). Subtract 7 from 19 to get 12. 12 isn't a perfect square, so simplify the root. You should end up with 2 root 3, aka
2 \sqrt{3}

For question 4 you must find the sqared value of a, b, and c. The sum of the sqared values of a and b is equal to the squared value of c. If c is 10 and a and b are both x, then the formula is
{x}^{2} +  {x}^{2} = 100
or
2 {x}^{2} = 100
Next, simplify this by first dividing by 2 on each side. You are then left with x squared equals 50. Simplify the root to 5 root 2, aka
5 \sqrt{2}

For question 5 you need to find the squared value of 47 and 25. This leaves you a hypotenuse of 2209 and a width of 625. Subtract 625 from 2209. This leaves you with 1584. Find the sqare root of this to find your length. Since it isn't a perfect square, simplify the root to get 12 root 11, aka
12 \sqrt{11}

I hope this helps clear things up for you!
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How would I do this problem?
kenny6666 [7]

If it's a geometric sequence then:

a_1=27;\ a_2=27\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{27}{36}=\dfrac{3}{4}=0.75\\\\a_{n+1}=a_nr\\\\a_3=27\cdot0.75=20.25\ CORRECT

We calculate the fourth and fifth term of the sequence:

a_4=a_3r\to a_4=20.25\cdot0.75=15.1875\\\\a_5=a_4r\to a_5=15.1875\cdot0.75=11.390625

Answer:

In year 4 15.1875 animals.

In year 5 11.390625 animals.

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3 years ago
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This seems like a right triangle problem.

So assuming 14 feet is the hypotenuse and 4 feet is a leg of the right triangle, we can use the pythagorean theorem (a^{2} +b^{2} =c^{2}) to solve for the height of the house, in which we shall name it x.

So, the equation is 4^{2} +x^{2} =14^{2}.

Solve for x:

16+x^{2}= 196

x^{2}= 196-16

x^{2}= 180

x = 6√5 feet


Hope this helps!

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