The answer would be (2,1)
Step-by-step explanation:
option is the correct answer the weight in X weeks of puppy that game 2 Pounds per week if it is starting weight is 8 pounds
The evaluate expression is 8
Given
<em>e</em> ˣʸ = sec(<em>x</em> ²)
take the derivative of both sides:
d/d<em>x</em> [<em>e</em> ˣʸ] = d/d<em>x</em> [sec(<em>x</em> ²)]
Use the chain rule:
<em>e</em> ˣʸ d/d<em>x</em> [<em>xy</em>] = sec(<em>x</em> ²) tan(<em>x</em> ²) d/d<em>x</em> [<em>x</em> ²]
Use the product rule on the left, and the power rule on the right:
<em>e</em> ˣʸ (<em>x</em> d<em>y</em>/d<em>x</em> + <em>y</em>) = sec(<em>x</em> ²) tan(<em>x</em> ²) (2<em>x</em>)
Solve for d<em>y</em>/d<em>x</em> :
<em>e</em> ˣʸ (<em>x</em> d<em>y</em>/d<em>x</em> + <em>y</em>) = 2<em>x</em> sec(<em>x</em> ²) tan(<em>x</em> ²)
<em>x</em> d<em>y</em>/d<em>x</em> + <em>y</em> = 2<em>x</em> <em>e</em> ⁻ˣʸ sec(<em>x</em> ²) tan(<em>x</em> ²)
<em>x</em> d<em>y</em>/d<em>x</em> = 2<em>x</em> <em>e</em> ⁻ˣʸ sec(<em>x</em> ²) tan(<em>x</em> ²) - <em>y</em>
d<em>y</em>/d<em>x</em> = 2<em>e</em> ⁻ˣʸ sec(<em>x</em> ²) tan(<em>x</em> ²) - <em>y</em>/<em>x</em>
Since <em>e</em> ˣʸ = sec(<em>x</em> ²), we simplify further to get
d<em>y</em>/d<em>x</em> = 2 tan(<em>x</em> ²) - <em>y</em>/<em>x</em>
Answer:
option (b) 700 mg to 740 mg
Step-by-step explanation:
Data provided;
The weight of candies is approximately normally distributed with mean 720 milligrams (mg)
this means that the most of the candies weight lies around the 720 milligrams
i.e
the most of the weight will be distributed as:
Mean ± standard deviation
here, the mean is 720 milligrams
thus,
the weights will lie within the range of some standard deviation above and below the mean and this is justified by the option (b) 700 mg to 740 mg
Hence,
the correct answer is option (b) 700 mg to 740 mg