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Annie M.
Tutor for 3 Years, College Student
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Pre-Algebra
TutorMe
Question:

George has to find a box that can fit 264 cubic cm of sand. The base of the box must be at least 11 cm long and 4 cm wide. How high should the box be to fit all the sand?

Annie M.
Answer:

We know the volume of the box must equal 264 cubic cm. The equation for volume: V = L x W x H Were solving for V. We know: V=264 cubic cm L= 11 cm W= 4 cm So when we plug the known variables into our equation we get: 264 cubic cm = 11 cm x 4 cm x H Now we must isolate H to find the answer: Divide both sides by 11 cm: 24 cm squared = 4 cm x H Divide both sides by 4 cm: 6 cm = H So the box should be 6 cm high.

Calculus
TutorMe
Question:

Let f(x) = $$ \frac{3x^{2}+7}{5x^{4}} $$ Find $$ \frac{d}{dx}. (f') $$(the derivative of x)

Annie M.
Answer:

We will use the quotient rule: $$ \frac{d}{dx} $$[$$ \frac{N(x)}{D(x)} $$] = $$ \frac{N'(x)D(x) - D'(x)N(x)}{D(x)^2} $$ N = Numerator D = Denominator. So lets write out the info we will put into our equation piece by piece: N = 3x^{2}+7 N' = 6x D = 5x^{4} D' = 20x^{3} Now we will plug this information into the quotient rule to get: $$ \frac{d}{dx}$$= $$ \frac{(6x)(5x^{4}) - (20x^{3})(3x^{2}+7)}{({5x^{4}})^2} $$ Lastly, simplify to get: $$ \frac{d}{dx}$$= $$ \frac{(30x^{5}) - (60x^{5}+140x^{3})}{({25x^{8}})} $$ = $$ \frac{-30x^{5} -140x^{3}}{{25x^{8}}} $$ = $$ \frac{-6x^{2} -28}{{5x^{5}}} $$

Algebra
TutorMe
Question:

Susan plans to jog 6 miles on the treadmill at a steady rate of 5 miles per hour. After jogging for 40 minutes, she wants to know how many more minutes it will take her to complete her goal distance of 6 miles.

Annie M.
Answer:

***Step 1: What do we know? Goal Distance = 6 miles Rate = 5 miles per hour Time Used = 40 minutes ***Step 2: Convert the rate into miles per minutes 1 hour = 60 minutes $$ \frac{5 miles}{1 hour} $$ = $$ \frac{5 miles}{60 minutes} $$ = $$ \frac{1 mile}{12 minutes} $$ So we know it takes her 12 minutes to run 1 mile. ***Step 3: Find how many minutes total it will take her to jog 6 miles. $$ \frac{1 mile}{12 minutes} $$=$$ \frac{6 miles}{X minutes} $$ Cross multiply: (1)(X) = (6)(12), thus X=72 So we know it will take her a total of 72 minutes to run the full 6 miles. ***Step 4: Find how many minutes she still has to jog: Total time - Time used = Time remaining 72 minutes - 40 minutes = 32 minutes remaining Therefore, to complete her goal of 6 miles, Susan will be jogging for 32 more minutes.

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