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Dafna11 [192]
3 years ago
7

Can I have the answer (15 points)​

Mathematics
1 answer:
hodyreva [135]3 years ago
4 0
N=9..................
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Please help me. if you can help me answer multiple questions only 3. offering 30 points for it
lubasha [3.4K]

Answer:

(-9, -5)

Step-by-step explanation:

Ok, so when you move an image to the right, you are moving along the x-axis, and when you move up, you are moving up the y-axis. So if the altered image is (x,y) and the values are (-5, -1), you reverse what has been done to the image. In this case, since we moved to the right 4 units, we know that means we added 4 to x, so we subtract 4 to get -9. And then, for the y-value, because we added 4, we do the opposite, and subtract 4 to get -5. So the pre-image should be (-9, -5)

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3 years ago
HailyDenney2 I need your assistance please!
ExtremeBDS [4]

Answer:

60 x 35 = 2,100

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Step-by-step explanation:

3 0
3 years ago
What is the area of the polygon? Show your work! ​
stepan [7]

Answer:

6+6=12 cm

4+4=8 cm

6+6= 12 cm

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Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
(3x + 1/2 ) + ( 7x - 4 1/2 ) simply the expression
ololo11 [35]

(3x + 1/2 ) + ( 7x - 4 1/2 )

3x+7x + 1/2 - 4 1/2

10x -4

5 0
3 years ago
(5x^3-8x^2+9x+12)/(x-3)<br><br> 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}

6 0
2 years ago
Read 2 more answers
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