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FrozenT [24]
3 years ago
15

What the answer if number 8

Mathematics
2 answers:
Alenkasestr [34]3 years ago
4 0

Answer: 2.93 lb


Step-by-step explanation:

1. To solve this problem you can rewrite the pounds given in the problem as a simple fractions:

6^{\frac{1}{8}}=\frac{(6*8)+1}{8}=\frac{49}{8}

 9^{\frac{1}{16}}=\frac{(9*16)+1}{16}=\frac{145}{16}

2. Now, you must subtract both weights to calculate the difference, as following. Then, you have the result shown below:

\frac{145}{16}lb-\frac{49}{8}lb=\frac{47}{16}lb=2.93lb

jeyben [28]3 years ago
3 0

<u>Answer:</u>

2.9375 pounds or 2\frac{15}{16} pounds.

<u>Step-by-step explanation:</u>

We know that Pedro's puppy weighed 6\frac{1}{8} pounds in his first vet visit while on his second visit, he weighed 9\frac{1}{16} pounds and we to find his weight gain.

To find the weight gained by Pedro's puppy, we simply need to find the difference of his two weights.

Converting mixed numbers into decimal number and then taking their difference out:

6\frac{1}{8} = 6.125

9\frac{1}{16} = 9.0625

Pedro's weight gain = 9.0625 - 6.125 = 2.9375 pounds or 2\frac{15}{16} pounds.

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If someone could help me out with this, I would greatly appreciate it!!
kati45 [8]

We are given that revenue of Tacos is given by the mathematical expression -7x^{2}+32x+240.

(A) The constant term in this revenue function is 240 and it represents the revenue when price per Taco is $4. That is, 240 dollars is the revenue without making any incremental increase in the price.

(B) Let us factor the given revenue expression.

-7x^{2}+32x+240=-7x^{2}+60x-28x+240\\-7x^{2}+32x+240=x(-7x+60)+4(-7x+60)\\-7x^{2}+32x+240=(-7x+60)(x+4)\\

Therefore, correct option for part (B) is the third option.

(C) The factor (-7x+60) represents the number of Tacos sold per day after increasing the price x times. Factor (4+x) represents the new price after making x increments of 1 dollar.

(D) Writing the polynomial in factored form gives us the expression for new price as well as the expression for number of Tacos sold per day after making x increments of 1 dollar to the price.

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7 0
3 years ago
For fun question
ZanzabumX [31]
Consider the function f(x)=x^{1/3}, which has derivative f'(x)=\dfrac13x^{-2/3}.

The linear approximation of f(x) for some value x within a neighborhood of x=c is given by

f(x)\approx f'(c)(x-c)+f(c)

Let c=64. Then (63.97)^{1/3} can be estimated to be

f(63.97)\approxf'(64)(63.97-64)+f(64)
\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375

Since f'(x)>0 for x>0, it follows that f(x) must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function f(x). This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...
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Answer: 15

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7 0
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