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2 years ago

WILL GIVE BRAINLIEST, ANSWER FAST PLEASE

Mathematics

1 answer:

Marizza181 [45]2 years ago

6 0

Answer:

$2( \frac{2}{5} x + 2) = \frac{4}{5} x + 4$

Step-by-step explanation:

Use the distributive property.

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Answer:

y + 2 = (-4/5)(x - 3)

Step-by-step explanation:

Your slope is -4/5. Your line passes through (3, -2).

Applying the point slope formula, we get:

y + 2 = (-4/5)(x - 3)

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3 years ago

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3 years ago

If 30 grams of coffee beans are used to make 48 fluid ounces how many grams are used for 100 fluid ounces

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3 years ago

About how many times greater is the volume of the cylinder than the volume of the cone?

Mamont248 [21]

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3 years ago

Find the absolute maximum and absolute minimum values of f on the given interval.

anyanavicka [17]

The question is missing parts. Here is the complete question.

Find the absolute maximum and absolute minimum values of f on the given interval.

$f(x)=xe^{-\frac{x^{2}}{32} }$ , [ -2,8]

Answer: Absolute maximum: f(4) = 2.42;

Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

Absolute maximum is a point where the function has its greatest possible value.

Absolute minimum is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

$f(x)=xe^{-\frac{x^{2}}{32} }$

Using the product rule, first derivative will be:

$f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )$

$f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )$ = 0

$1-\frac{x^{2}}{16}=0$

$\frac{x^{2}}{16}=1$

$x^{2}=16$

x = ±4

x can't be -4 because it is not in the interval [-2,8].

$f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76$

$f(4)=4e^{-\frac{4^{2}}{32} }=2.42$

$f(8)=8e^{-\frac{8^{2}}{32} }=1.08$

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

8 0

3 years ago