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3 years ago

I need to simplify this complex fraction

Mathematics

1 answer:

vovangra [49]3 years ago

7 0

I see two ways to do it.

First, you have to understand that when you see a 'complex fraction' like this, the top number is a numerator, and the two bottom numbers are both denominators.

Way-1:

Take the top fraction . . . 2/2 . That's equal to 1 . So the whole thing is 1/3.

Way-2:

Multiply the bottom two denominators. Then the whole thing is 2/(2·3) . That's the same thing as 2/6 . Simplify that, and you have 1/3 .

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A radius of 4.1 yards. What is the area? Use 3.14 for. yards

Pani-rosa [81]

Answer:

52.7834

Step-by-step explanation:

7 0

2 years ago

23 and 25 how do you do the work? find the areas of each

stiks02 [169]

9514 1404 393

Answer:

23) 35.77 in²

25) 48.19 cm²

Step-by-step explanation:

Use the appropriate area formula with the given information.

23) The area of a triangle is given by the formula ...

A = 1/2bh . . . . . base b, height h

A = 1/2(9.8 in)(7.3 in) = 35.77 in²

25) The area of a parallelogram is given by the formula ...

A = bh . . . . . . base b, height h

A = (7.9 cm)(6.1 cm) = 48.19 cm²

_____

The height in each figure is measured perpendicular to the base. This tells you that the length 10.6 cm of the diagonal side of the triangle is not relevant to finding the area.

6 0

3 years ago

Read 2 more answers

Help appreciated on question in image! Thanks:)

Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

$x^4+8x^3+8x^2+8x+7=0$

Solving

$x^4+8x^3+8x^2+8x+7$

$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=7,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$ $\mathrm{Solve\:}\:x+1=0:\quad x=-1$