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2 years ago

Some one please help

Mathematics

1 answer:

chubhunter [2.5K]2 years ago

6 0

Answer:

Its C

Step-by-step explanation:

Rational numbers contain decimals that do not repeat and whole numbers

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. Which ordered pairs in the form (x, y) are solutions to the equation 7x – 5y = 28?

Brut [27]

we know that

If a ordered pair (x,y) is a solution of the equation, then the ordered pair must satisfy the equation

we have the equation

$7x-5y=28$

Let's verify all the cases to determine the solution to the problem.

case A) point $(-6,-14)$

Substitute the values of x and y in the equation

$x=-6\\y=-14$

$7(-6)-5(-14)=28$

$-42+70=28$

$28=28$ -------> is true

therefore

The point $(-6,-14)$ is a solution of the equation

case B) point $(-1,-7)$

Substitute the values of x and y in the equation

$x=-1\\y=-7$

$7(-1)-5(-7)=28$

$-7+35=28$

$28=28$ -------> is true

therefore

The point $(-1,-7)$ is a solution of the equation

case C) point $(4,10)$

Substitute the values of x and y in the equation

$x=4\\y=10$

$7(4)-5(10)=28$

$28-50=28$

$-22=28$ -------> is not true

therefore

The point $(4,10)$ is not a solution of the equation

case D) point $(7,9)$

Substitute the values of x and y in the equation

$x=7\\y=9$

$7(7)-5(9)=28$

$49-45=28$

$4=28$ -------> is not true

therefore

The point $(7,9)$ is not a solution of the equation

therefore

the answer is

$(-6,-14)$

$(-1,-7)$

8 0

3 years ago

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I need help on both questions!

ankoles [38]

Answer:

9 sq.m

Step-by-step explanation:

because the area of triangle is 1/2 *b*h

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3 years ago

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Does the Commutative Property of Addition apply when you add two negative integers?

MAVERICK [17]

The commutative property of addition means we can add two integers in any order. So yes, It would still apply to two negative integers (for example, -2 + -3 and -3 + -2 both equal five)

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3 years ago

Need help ASAP due in 30 minutes!

Whitepunk [10]

Y=3x-4

Hope you get it turned in in time!!!

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3 years ago

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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

2 years ago