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4 years ago

5

Which expression is equivalent to 84+48

1 answer:

padilas [110]4 years ago

4 0

The answer is a
-hope this helped-

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How do you solve for the value of PI using the formula for circumference?

stira [4]

$\text{The circumference of a circle}\\\\C=d\pi\\\\C-circumference\\d-diameter\\\\C=d\pi\to\pi=\dfrac{C}{d}\\------------------------------\\\\\text{The area of a circle}\\\\A=\pi r^2\\\\A-area\\r-radius\\\\A=\pi r^2\to\pi=\dfrac{A}{r^2}$

$EXAMPLE:\\\\d=7\ in\to2d=7\ in\to d=3.5\ in\\C=21.98\ in\\A=38.465\ in^2\\\\\pi=\dfrac{C}{d}\to\pi=\dfrac{21.98}{7}=3.14\\\\\pi=\dfrac{A}{r^2}\to\pi=\dfrac{38.465}{3.5^2}=\dfrac{38.465}{12.25}=3.14$

6 0

4 years ago

Plz help me ASAP thx.

kolbaska11 [484]

Use the Pythagorean theorem:

$a^2+(2\sqrt3)^2=(4\sqrt3)^2\\\\a^2+2^2(\sqrt3)^2=4^2(\sqrt3)^2\\\\a^2+4\cdot3=16\cdot3\\\\a^2+12=48\ \ \ \ |-12\\\\a^2=36\to a=\sqrt{36}\\\\\boxed{a=6\to C.}$

Used:

$(a\cdot b)^n=a^n\cdot b^n\\\\(\sqrt{a})^2=a$

3 0

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.

<u>Absolute</u> <u>maximum</u> is a point where the function has its greatest possible value.

<u>Absolute</u> <u>minimum</u> 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

Which expression is equivalent to<br> (4x^3y^5)(3x^5y)^2

Ivahew [28]

Answer:

que es eso no se ingles

8 0

2 years ago

Please help me on this will give you brainliest

erik [133]

X would be 17 or 17.32

7 0

3 years ago

Read 2 more answers