<u>Answer:</u>
<u>Step-by-step explanation:</u>
<u>Let's find 'a' using Pythagoras theorem.</u>
- => 4² = 2² + a²
- => 16 = 4 + a²
- => 12 = a²
- => a = √12
- => a = √2 x 2 x 3
- => a = 2√3
- => a = 2 x √3
- => a = 2 x 1.732
- => a = 3.464 = 3.5 (Estimated)
Hoped this helped.

Answer:
x=3.4
y=-6.75
Step-by-step explanation:
5x+4y=-10 (equation 1 )
-5x+3y=-17 (equation 2)
7y=-27
y=-6.75
substitution (equation 1 ) y=-6.75
5x-27=-10
5x=17
x=3.4
Answer:
13 and 11
Step-by-step explanation:
how is helpful
Answer:
9
Step-by-step explanation:
(15×4+3)/4 ÷(4×1+3)/4= 63/4 ÷ 7/4= 9
Answer: ![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Step-by-step explanation:

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;


Divide by m to isolate
.


To eliminate the square and isolate v, extract the square root.
![\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3D%5Csqrt%5B%5D%7Bv%5E2%7D)
![\sqrt[]{\frac{2K}{m} }=v](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3Dv)
let's rewrite it in a way that v is in the left side.
![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)