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3 years ago

In order to answer the question correctly, please use the image below:

Mathematics

2 answers:

Maurinko [17]3 years ago

5 0

Answer:

240

Step-by-step explanation:

m xyz 2(120)

(sorry about the explaination)

saul85 [17]3 years ago

5 0

Answer:

The answer to this is 240.

Step-by-step explanation:

m xyz 2(120)=240

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Solve the rational function of (check image) and check for extraneous solutions

Oxana [17]

Answer:

Option A is correct.

i.e. x = 1, x = 0 is an extraneous solution.

Step-by-step explanation:

Given the expression

$\frac{5}{x}=\frac{4x+1}{x^2}$

Solving the rational function

$\frac{5}{x}=\frac{4x+1}{x^2}$

Apply fraction across multiply: if $\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$5x^2=x\left(4x+1\right)$

Subtract x(4x+1) from both sides

$5x^2-x\left(4x+1\right)=x\left(4x+1\right)-x\left(4x+1\right)$

Simplify

$5x^2-x\left(4x+1\right)=0$

5x² - 4x² - x = 0

x² - x = 0

Factor x² - x = x(x-1)

x(x-1) = 0

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

$x=0\quad \mathrm{or}\quad \:x-1=0$

Thus, the solution to the equation is:

$x=0,\:x=1$

But, it is clear that if we substitute x = 0, the equation becomes undefined because we can not have the denominator to be 0.

In other words, the equation is undefined for x = 0

Thus, x = 0 is an extraneous solutions.

Therefore, option A is correct.

i.e. x = 1, x = 0 is an extraneous solution.

5 0

3 years ago

Evaluate exactly the value of the integral from -1 to 0 (2x^4+8x)^3 (4x^3+4),dx. Include the use of substitution and the antider

ddd [48]

Hello,

Let's be

$z=2x^4+8x \\
dz=(8x^3+8)\ dx=8(x^3+1)\ dx \\

 \int\limits^0_{-1} {(2x^4+8x)^3*(4x^3+4)} \, dx =4* \int\limits^0_{-6} { \dfrac{z^3}{2}} \, dz \\
= 2* \left[\begin{array}{ccc} \dfrac{z^4}{4}\end{array}\right] ^0_{-6}=
\left[\begin{array}{ccc} \dfrac{z^4}{2}\end{array}\right] ^0_{-6}=0-648=-648$

3 0

4 years ago

What is the midpoint of a segment with These endpoints (-12,-3) and (3,-8)

Savatey [412]

Answer:

$(\frac{-9}{2},\frac{11}{2})$

Step-by-step explanation:

you have to use the midpoint formula which is

$(\frac{x1+x2}{2},\frac{y1+y2}{2})$

so you plug your numbers in

$(\frac{-12+3}{2},\frac{-3+-8}{2})$

Add the tops

$(\frac{-9}{2} ,\frac{-11}{2} )$

if possible simplify

and that is your answer

8 0

2 years ago

Whoever answers these 2 right gets brainliest and 25 points!

Lana71 [14]

The first one is B and the second one is C

3 0

3 years ago

Describe at least two differences between constructing an inscribed regular hexagon and constructing an inscribed square.

liberstina [14]

Correct answer is A.

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.

<h3>How to calculate the diameter?</h3>

If you know the radius of the circle, multiply by 2 to get the diameter. The radius is the distance from the center of the circle to the edge. For example, if the radius of a circle is 4 cm, then the diameter is 4 cm x 2 = 8 cm. If you know the circumference of the circle, divide by π to get the diameter.

<h3>Briefing:</h3>

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction which is A.

The diameter of the circle is used for the square as l=2r therefore corresponds to the diameter, while for a hexagon r=r being the radius of the circle.

To know more about Diameter visit:

brainly.com/question/5501950

#SPJ4

I understand that the question you are looking for is:

Describe at least two differences between constructing an inscribed regular hexagon and constructing an inscribed square.

A) The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.

B) The radius of the circle is used for the square construction, but the diameter of the circle is used for the regular hexagon construction.

C) The square will need two arcs along the circle, but the hexagon will need two arcs above and two arcs below the diameter of the circle.

D) The square will need six arcs along the circle, but the hexagon will need two arcs above and two arcs below the diameter of the circle.

7 0

2 years ago