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3 years ago

Find the Surface Area of this Prism

Mathematics

1 answer:

lyudmila [28]3 years ago

5 0

Answer:

$54 m^2$ is the correct answer I believe

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Given that the two triangles shown are congruent, explain one way to verify that the corresponding angles of the two triangles a

dimaraw [331]

Answer:

A) Is correct

Step-by-step explanation:

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3 years ago

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Determine if the expression -5y^5-y^3−5y5−y3 is a polynomial or not. If it is a polynomial, state the type and degree of the pol

8_murik_8 [283]

Answer:I think it y=73

Step-by-step explanation:

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4 years ago

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If a rectangle measures 20 meters by 48 meters, what is the length, in meters, of the rectangle?

miv72 [106K]

P of a rectangle= 2l + 2w

P = 2(48) + 2(20)
P= 96 + 40
P= 136m

The length of the rectangle is 136 meters.

I hope that helped.

6 0

4 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

Write an equation of the line below. HELP PLEASE!!

timurjin [86]

Two points: (-5,7), (0,2)
Find slope: (2-7)/(0+5) = -5/5 = -1
Y intercept is given: 2
Y = -1x + 2

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3 years ago