l = 2w - 4
Because we're solving for 2l + 2w, that can be simplified to
2(2w - 4) + 2w = 34
4w - 8 + 2w = 34
6w - 8 = 34
6w = 42
w = 7
Knowing this, we can input w:
2(7) + 2l = 34
14 + 2L = 34
2l = 20
l = 10
L = 10, W = 7, Option C
Answer:
The cosine of the angle is 4/5.
Step-by-step explanation:
An acute angle is between 0º and 90º, in the first quadrant of the trigonometric circle, in which both the sine and the cosine are positive values.
For each angle
, we have that:
![\sin^{2}{\alpha} + \cos^{2}{\alpha} = 1](https://tex.z-dn.net/?f=%5Csin%5E%7B2%7D%7B%5Calpha%7D%20%2B%20%5Ccos%5E%7B2%7D%7B%5Calpha%7D%20%3D%201)
We have that:
![\sin{\alpha} = \frac{3}{5}](https://tex.z-dn.net/?f=%5Csin%7B%5Calpha%7D%20%3D%20%5Cfrac%7B3%7D%7B5%7D)
So
![\sin^{2}{\alpha} + \cos^{2}{\alpha} = 1](https://tex.z-dn.net/?f=%5Csin%5E%7B2%7D%7B%5Calpha%7D%20%2B%20%5Ccos%5E%7B2%7D%7B%5Calpha%7D%20%3D%201)
![(\frac{3}{5})^{2} + \cos^{2}{\alpha} = 1](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B5%7D%29%5E%7B2%7D%20%2B%20%5Ccos%5E%7B2%7D%7B%5Calpha%7D%20%3D%201)
![\cos^{2}{\alpha} = \frac{16}{25}](https://tex.z-dn.net/?f=%5Ccos%5E%7B2%7D%7B%5Calpha%7D%20%3D%20%5Cfrac%7B16%7D%7B25%7D)
![\cos{\alpha} = \frac{4}{5}](https://tex.z-dn.net/?f=%5Ccos%7B%5Calpha%7D%20%3D%20%5Cfrac%7B4%7D%7B5%7D)
The cosine of the angle is 4/5.
1. given
2. addition property of equality
3. division property of equality
Answer:
The value of the ve = 9m/sec
Step-by-step explanation:
From the given formula, it can be conclude that this is evenly accelerated movement.
First I will rewrite given formula
vi = √ ve∧2 - 2ad First we will square on both sides and get
vi∧2 = ve∧2 - 2ad Now we will add monom (+2ad) to both sides and get
vi∧2 + 2ad = ve∧2 -2ad + 2ad =>
ve∧2 = vi∧2 +2ad Now we will rooted both sides and get
ve = √vi∧2 + 2ad
Now we will replace given data vi=7m/sec, a=8m/s∧2 and d=2m in the last formula
ve= √7∧2 + 2*8*2 = √49+32 = √81 = 9
ve= 9m/sec
Good luck!!!