Answer:
0.15-0.75p? YES
-0.45-0.6+0.75P? NO
-3(-0.05+0.25p)? YES
-3(0.15)+(-3)(0.2)+(-3)(0.25p)? NO
Step-by-step explanation:
-3(0.15-0.2+0.25p)
=-0.45+0.6-0.75p
=0.15-0.75p
1. 0.15-0.75p=0.15-0.75p? YES
2. -0.45-0.6+0.75P=0.15-0.75p?
-1.05+0.75p=0.15-0.75p? NO
3. -3(-0.05+0.25p)=0.15-0.75p?
0.15-0.75=0.15-0.75p? YES
4. -3(0.15)+(-3)(0.2)+(-3)(0.25p)=0.15-0.75p?
-0.45-0.6-0.75p=0.15-0.75p?
-1.05-0.75p=0.15-0.75p? NO
X³ = -1000
x = ∛-1000
x = -10
In short, Your Answer would be: -10
Hope this helps!
Answer:
∠WUV
Step-by-step explanation:
Start from the top, the meeting point of the rays is the angle U
Answer:
y = 3x-6
Step-by-step explanation:
y = 3x−5
This is in slope intercept form
y = mx+b where m is the slope and b is the y intercept
m =3
Parallel lines have the same slope
We have the slope m=3 and a point (2,0)
y = mx+b
y = 3x+b
Substituting the point into the equation to solve for b
0 = 3(2)+b
0 = 6+b
b = -6
y = 3x-6
well, let's first notice, all our dimensions or measures must be using the same unit, so could convert the height to liters or the liters to centimeters, well hmm let's convert the volume of 1000 litres to cubic centimeters, keeping in mind that there are 1000 cm³ in 1 litre.
well, 1000 * 1000 = 1,000,000 cm³, so that's 1000 litres.
![\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=1000000~cm^3\\ h=224~cm \end{cases}\implies \stackrel{cm^3}{1000000}=\pi r^2(\stackrel{cm}{224}) \\\\\\ \cfrac{1000000}{224\pi }=r^2\implies \sqrt{\cfrac{1000000}{224\pi }}=r\implies \cfrac{1000}{\sqrt{224\pi }}=r\implies \stackrel{cm}{37.7}\approx r](https://tex.z-dn.net/?f=%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%5C%5C%5C%5C%20V%3D%5Cpi%20r%5E2%20h~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D1000000~cm%5E3%5C%5C%20h%3D224~cm%20%5Cend%7Bcases%7D%5Cimplies%20%5Cstackrel%7Bcm%5E3%7D%7B1000000%7D%3D%5Cpi%20r%5E2%28%5Cstackrel%7Bcm%7D%7B224%7D%29%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B1000000%7D%7B224%5Cpi%20%7D%3Dr%5E2%5Cimplies%20%5Csqrt%7B%5Ccfrac%7B1000000%7D%7B224%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Ccfrac%7B1000%7D%7B%5Csqrt%7B224%5Cpi%20%7D%7D%3Dr%5Cimplies%20%5Cstackrel%7Bcm%7D%7B37.7%7D%5Capprox%20r)
now, we could have included the "cm³ and cm" units for the volume as well as the height in the calculations, and their simplication will have been just the "cm" anyway.