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3 years ago

Help pls :) question is in attachment

Mathematics

1 answer:

nekit [7.7K]3 years ago

6 0

Answer:

x = 1

Step-by-step explanation:

because that's the point where the two lines intersect!

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Identify the Domain and Range of the graph below: Domain: Range:

Nuetrik [128]

Domain is all real numbers
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3 years ago

How to fit 6 into 10 more than once so you get a remander

zhannawk [14.2K]

Answer:

It can go in once with a remainder of 4

Step-by-step explanation:

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4 years ago

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Neil has 3 partially full cans of white paint. They cantain 1/3 gallon, 1/5 gallon, and 1/2 gallon of paint. About how much pain

Marianna [84]

Answer:

1 1/30 gallons, or 31/30 gallons of paint

Step-by-step explanation:

Before these fractions can be added together they have to have a common denominator. The Least Common Multiple of 2, 3, and 5 is 30, so

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3 years ago

GEOMETRY PEOPLE PLEASE HELP ME!!!!!!!!!!!!!!!!! I WILL GIVE BRAINLIEST!

IrinaVladis [17]

$\Delta KLJ\sim\DeltaNPM$ (AAA) Corresponding angles are congruent.

Therefore, the sides of the triangles are proportional:

$\dfrac{4x-4}{20}=\dfrac{30}{25}$ cross multiply

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--------------------------------------------------------------------------------------------

$\Delta PQR\cong\Delta STU$ (SAS)

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

Therefore we have the equation:

$3y-2=y+4\qquad|+2\\\\3y=y+6\qquad|-y\\\\2y=6\qquad|:2\\\\y=3$

The perimeter of △PQR:

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3 years ago

(5x^3-8x^2+9x+12)/(x-3) please answer with sentences on how to do it step by step. thanks!

Anastaziya [24]

Answer:

$5x^2+7x+30+\frac{102}{x-3}$

Step-by-step explanation:

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}5x^3-8x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{5x^3}{x}=5x^2\\\mathrm{Quotient}=5x^2\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}5x^2:\:5x^3-15x^2\\\mathrm{Subtract\:}5x^3-15x^2\mathrm{\:from\:}5x^3-8x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=7x^2+9x+12\\\mathrm{Therefore}\\=5x^2+\frac{7x^2+9x+12}{x-3}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}7x^2+9x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{7x^2}{x}=7x\\\mathrm{Quotient}=7x\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}7x:\:7x^2-21x\\\mathrm{Subtract\:}7x^2-21x\mathrm{\:from\:}7x^2+9x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=30x+12\\\mathrm{Therefore}\\=5x^2+7x+\frac{30x+12}{x-3}\\$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}30x+12\\\mathrm{and\:the\:divisor\:}x-3\mathrm{\::\:}\frac{30x}{x}=30\\\mathrm{Quotient}=30\\\mathrm{Multiply\:}x-3\mathrm{\:by\:}30:\:30x-90\\\mathrm{Subtract\:}30x-90\mathrm{\:from\:}30x+12\mathrm{\:to\:get\:new\:remainder}\\\mathrm{Remainder}=102\\\mathrm{Therefore}\\=5x^2+7x+30+\frac{102}{x-3}$

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3 years ago

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