Hope this helps but I think it might be:
Part A: c=0.75p
the cost of shipping for 2 pounds
c=0.75p
c=0.75*2
c=$1.50
Part B: c=0.05p
Answer:
<h3>
- 128 Superscript StartFraction 3 Over x EndFraction
</h3><h3>
- (4RootIndex 3 StartRoot 2 EndRoot)x
</h3><h3>
- (4 (2 Superscript one-third Baseline) ) Superscript x</h3><h3>
Step-by-step explanation:</h3>
Given the indicinal equation ![(\sqrt[3]{128} )^{x}\\](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B128%7D%20%29%5E%7Bx%7D%5C%5C)
According to one of the law of indices,
![(\sqrt[a]{m} )^{b}\\= (\sqrt{m})^\frac{b}{a}](https://tex.z-dn.net/?f=%28%5Csqrt%5Ba%5D%7Bm%7D%20%29%5E%7Bb%7D%5C%5C%3D%20%28%5Csqrt%7Bm%7D%29%5E%5Cfrac%7Bb%7D%7Ba%7D)
Applying this law to the question;
![(\sqrt[3]{128} )^{x}\\ = {128} ^\frac{x}{3}\\ \\= (\sqrt[3]{64*2})^{x} \\ = (4\sqrt[3]{2})^{x} \\= (4(2^{1/3} )^{x} )](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B128%7D%20%29%5E%7Bx%7D%5C%5C%20%3D%20%7B128%7D%20%5E%5Cfrac%7Bx%7D%7B3%7D%5C%5C%20%5C%5C%3D%20%28%5Csqrt%5B3%5D%7B64%2A2%7D%29%5E%7Bx%7D%20%5C%5C%20%3D%20%284%5Csqrt%5B3%5D%7B2%7D%29%5E%7Bx%7D%20%5C%5C%3D%20%284%282%5E%7B1%2F3%7D%20%29%5E%7Bx%7D%20%29)
The following are therefore true based on the following calculation
128 Superscript StartFraction 3 Over x EndFraction
(4RootIndex 3 StartRoot 2 EndRoot)x
(4 (2 Superscript one-third Baseline) ) Superscript x
The solution would be like this for this specific problem:


The correct sequence of operations when solving
negative one over five (x − 25) = 7 would be multiply each side by −5, and adding
25 to each side.
Answer:
48.6
Step-by-step explanation:
If you use 8.1g of sugar for 1 cake then 6 cakes will be 48.6g of sugar
Just do 8.1*6 and you will get 48.6
Since q+d=19, we can re-write this as d=19-q. Using the second equation 0.25q+0.1d=4 we can multiply both sides by 100. So we get 25q+10d=400. So now we can plug d=19-q into 25q+10d=400. So now we get, 25q+190-10q=400. Subtracting both sides by 190, we get 15q=210 and that q=14 plugging that in d=5