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3 years ago

Hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Mathematics

2 answers:

german3 years ago

7 0

9x9 = 81
Square root = 3
81-3 = 78

tatuchka [14]3 years ago

5 0

Hiiii? what is the question

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How to find the perimeter of a right triangle

QveST [7]

Answer:

There are three primary methods used to find the perimeter of a right triangle.

1. When side lengths are given, add them together.

2. Solve for a missing side using the Pythagorean theorem.

3. If we know side-angle-side information, solve for the missing side using the Law of Cosines.

Step-by-step explanation:

there i hope this helps!!!

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3 years ago

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Wewaii [24]

Answer:

can u be more specific so i can properly help u

Explanation:

pls explain and i will edit my answer

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3 years ago

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PLEASE HELP FAST The Booster Club raised $30,000 for a sports fund. No more money will be placed into the fund. Each year the fu

zloy xaker [14]

Answer:

c. $24,435

Step-by-step explanation:

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3 0

3 years ago

What is the area of a rectangle that has side lengths of 3/4 yard and 5/6 yard

Ahat [919]

We are asked to find the area of the rectangle that has side lengths of 3/4 yard and 5/6 yard.

We know that area of rectangle is width times length.

To find the area of the given rectangle we will multiply both side lengths as:

$\text{Area of rectangle}=\text{Width}\times \text{Length}$

$\text{Area of rectangle}=\frac{3}{4}\text{ yard}\times\frac{5}{6}\text{ yard}$

$\text{Area of rectangle}=\frac{3}{4}\times\frac{5}{6}\text{ yard}^2$

$\text{Area of rectangle}=\frac{1}{4}\times\frac{5}{2}\text{ yard}^2$

$\text{Area of rectangle}=\frac{1\times5}{4\times2}\text{ yard}^2$

$\text{Area of rectangle}=\frac{5}{8}\text{ yard}^2$

Therefore, the area of the given rectangle would be $\frac{5}{8}$ square yards.

5 0

3 years ago

What is the completely factored form of f(x)=x^3-7x^2+2x+4

xxMikexx [17]

$Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)$

$Steps:$

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

$Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}$

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

$\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}$

$\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4$

$\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

8 0

3 years ago