The answer is <span>She went over by one over six of a cup
</span>
<span>1 cup of that particular chocolate powder has a mass of 128 grams: 1c = 128g
</span><span>64 grams of chocolate powder is x cups.
1c = 128g
x = 64g
1c : 128 g = x : 64g
x = </span>1c : 128 g * 64 g
x = 0.5 c
64 grams of chocolate powder is 0.5 cups = 1/2 cups
<span>She added two over three of a cup of chocolate powder: 2/3 cups
She need to add 1/2 cups: 1/2 = 3/6 cups
She added 2/3 cups: 2/3 = 4/6 cups
So she added 4/6 - 3/6 = 1/6 cups more.
</span>
I'm guessing you want this written as an equation, so here it is.
3n+7>2(n-4)
These can be really tricky at times, so I suggest looking for "clues" in the wording, such as add, subtract, and other things. Also look for what numbers are "paired" together, like in parentheses. Hope that this helps!
Answer:
Step-by-step explanation:
B
Answer:
neither!
Step-by-step explanation:
That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
