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Katena32 [7]
3 years ago
9

Find two numbers whose difference is 102 and whose product is a minimum. Step 1 If two numbers have a difference of 102, and one

of them is x + 102, then the other is $$ Incorrect: Your answer is incorrect. x. Step 2 The product of two numbers x and x + 102 can be simplified to be x2 Correct: Your answer is correct. seenKey 2 + 102 Correct: Your answer is correct. seenKey 102 x. Step 3 If f(x) = x2 + 102x, then f '(x) = $$ Correct: Your answer is correct. 2x+102. Step 4 To minimize the product f(x) = x2 + 102x, we must solve 0 = f '(x) = 2x + 102, which means x = -51 Correct: Your answer is correct. seenKey -51 . Step 5 Since f ''(x) = 2 , there must be an absolute minimum at x = −51. Thus, the two numbers are as follows. (smaller number) (larger number)
Mathematics
1 answer:
NISA [10]3 years ago
4 0

Answer:

The two numbers would be -51 and 51

Step-by-step explanation:

To find these, first set the equation for the first number as x. You can then set the second number as x + 102. Now, find their product.

x(x + 102) = x^2 + 102x

Now, to find the minimum, find the value of x in the vertex of this equation.

-b/2a = -102/2(1) = -102/2 = -51

So we know -51 is the first number. Now we find the second using the prewritten equation.

x + 102 = -51 + 102 = 51

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7x + y = 13<br> 4x – 4y = -20
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Answer: (x,y) = (1,6)

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What is 3.3(x-8)-x=1.2
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Multiply 3.3 by everything in the parenthesis.

3.3x - 26.4 - x = 1.2

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2.3x - 26.4 = 1.2

Add 26.4 to both sides.

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3 years ago
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Can somebody please help me i need this urgently.
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multiple choice is worth 2

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Step-by-step explanation:

23x + 10y = 86

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-23x + -10y = -86

56x + 10y = 152

add them

33x = 66

x = 2

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6 0
2 years ago
A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dydt=8. What is x that this instant?
professor190 [17]

Answer:

The value of x at this instant is 3.

Step-by-step explanation:

Let x\cdot y = 18, we get an additional equation by implicit differentiation:

x\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0 (1)

From the first equation we find that:

x = \frac{18}{y} (2)

By applying (2) in (1), we get the resulting expression:

\frac{18}{y}\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0 (3)

y\cdot \frac{dx}{dt}=-\frac{18}{y}\cdot \frac{dy}{dt}

\frac{dx}{dt} = -\frac{18}{y^{2}} \cdot \frac{dy}{dt}

If we know that y = 6 and \frac{dy}{dt} = 8, then the first derivative of x in time is:

\frac{dx}{dt} = -\frac{18}{6^{2}} \cdot (8)

\frac{dx}{dt} = -4

From (1) we determine the value of x at this instant:

x\cdot \frac{dy}{dt} = -y\cdot \frac{dx}{dt}

x = -y\cdot \left(\frac{\frac{dx}{dt} }{\frac{dy}{dt} } \right)

x = -6\cdot \left(\frac{-4}{8} \right)

x = 3

The value of x at this instant is 3.

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3 years ago
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