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Diano4ka-milaya [45]
3 years ago
9

Linear Equation

Mathematics
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

See Below

Step-by-step explanation:

It's a lot easier to get a linear equation in slope-intercept form (y = mx + b, where m and b are the line's slope and y-intercept, respectively) if you want to graph it, so let's do that here.

Rewriting the given equation gives us:

5x - y = 12

-y = -5x + 12

y = 5x - 12

We see here that m = 5 and b = -12, so the line has a slope of 5 and a y-intercept of (0, -12). Using this information, we can graph the line, as shown below. Hope this helps!

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Part 1

6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)}  \\  \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\  \\ =3



Part 2

7-8\div4(3^2-1)=7-8\div4(9-1) \\  \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\  \\ =6  \frac{1}{2} = \frac{13}{2}



Part 3

2-2(2-5)^2+3^2=2-2(-3)^2+9 \\  \\ =2-2(9)+9=2-18+9=-7



Part 4

(5^2-9\cdot3)^2-11=(25-27)^2-11 \\  \\ =(-2)^2-11=4-11=-7



Part 5

[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\  \\ =(1-5)\cdot4=-4\cdot4=-16



Part 6

12-11\cdot2+16\div8=12-22+2 \\  \\ =-8



Part 7

-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\  \\ =-2-(-1)=-2+1=-1



Part 8

8+2\cdot3-14\div7=8+6-2 \\  \\ =12



Part 9

(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\  \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\  \\ =36+4=40



Part 10

7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\  \\ =7+(-7)\cdot5=7+(-35)=7-35 \\  \\ =-28



Part 11

3-2[2^3+(-1)^3]=3-2[8+(-1)] \\  \\ =3-2(8-1)=3-2(7)=3-14 \\  \\ -11



Part 12

18+6\div3(2^2+5)=18+6\div3(4+5) \\  \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\  \\  =18 \frac{2}{9}
8 0
4 years ago
Find the distance between (3,4) and (4,-6) if necessary, round to the nearest tenth.
Ratling [72]

Answer:

The distance is:

d = 10.0 units (Rounded to the nearest the Tenths Place)

Step-by-step explanation:

Given the points

  • (3,4)
  • (4,-6)

The distance 'd' between (3,4) and (4,-6)

\mathrm{Compute\:the\:distance\:between\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):

d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}

substituting the points values

   =\sqrt{\left(4-3\right)^2+\left(-6-4\right)^2}

   =\sqrt{1+10^2}

   =\sqrt{1+100}

   =\sqrt{101}

   =10.0  units (Rounded to the nearest the Tenths Place)

Thus, the distance is:

d = 10.0 units (Rounded to the nearest the Tenths Place)

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