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gayaneshka [121]
3 years ago
9

How do I solve this?

Mathematics
2 answers:
aksik [14]3 years ago
4 0
Multiply the first equation by 2
4x-6y=16

Then add the equations to eliminate x
4x-6y=16
-4x+5y=-10

-y=6
y-=6

Plug the y value in
2x-3(-6)=8
2x+18=8
2x=-10
x=-5

Final answer: (-5,-6)
Anastaziya [24]3 years ago
4 0
2x -3y = 8  --->  y = 2/3x +-8/3

-4x +5(2/3x + -8/3) = -10
-4x + 10/3x + -40/3 = -10
-2/3x + -40/3 = -10
plus 40/3 on both side = -2/3x = 10/3
multiply both side by 3/-2 = -5

x = -5      y = -6

y = sub x into orginal equation


 
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4 (b-6)+19 simplify the expression
BlackZzzverrR [31]
4b - 6b + 19
Answer : -2b + 19
3 0
3 years ago
Need some help with this question please.
Ksivusya [100]
Use Pythagorean theorem
a^2 + b^2 = c^2
Plug in the info
b = 53, a = 41
41^2 + 53^2 = c^2
1681 + 2809 = c^2
c^2 = 4490
Squareroot of 4490 = 67.007
Perimeter = adding all sides
53m + 41m + 67.01m = 161.01 m

Solution: 161.01 meters
8 0
2 years ago
PLS HELP ASAP The following probability distribution table shows information collected from a survey of students regarding their
gregori [183]

Answer:

44%

Step-by-step explanation:

For these types of problems, look at the totals. We know that there are 0.44 out of 1 eighth graders. Since it doesn't specify how the 8th grader is dropped off, we take 0.44 and divide it by 1 (the total). We get 0.44, but since it's most likely asking for percentages, multiply that by 100.

8 0
2 years ago
Please please please help
Eduardwww [97]

Answer:

I. B. true

II. A true

Step-by-step explanation:

we can use the quadratic equation:

x=\frac{-b+-\sqrt{b^{2}-4*a*c}}{2*a}

we have:

1)x^{2} -4x+5

a=1 b=-4 c=5

x=\frac{-(-4)+-\sqrt{(-4)^{2}-4*1*5}}{2*1} \\\\x=\frac{4+-\sqrt{-4}}{2} \\\\x=\frac{4+-\sqrt{4} i}{2}\\\\x=\frac{4+-2i}{2} \\\\x_{1} =2-i\\\\x_{2} =2+i

the quadratic expression has two complex factors.

2)x^{2} -5x+14

a=1 b=-5 c=14

x=\frac{-(-5)+-\sqrt{(-5)^{2}-4*1*14}}{2*1} \\\\x=\frac{5+-\sqrt{-31}}{2} \\\\x=\frac{5+-\sqrt{31} i}{2}\\\\x_{1} =\frac{5-\sqrt{31}i }{2}\\ \\x_{2} =\frac{5+\sqrt{31}i }{2}

the solution of x^{2} -5x+14 are

x_{1} =\frac{5-\sqrt{31}i }{2}\\\\or\\x_{2} =\frac{5+\sqrt{31}i }{2}

3)2x^{2} -8x+5

a=2 b=-8 c=5

x=\frac{-(-8)+-\sqrt{(-8)^{2}-4*2*5}}{2*2} \\\\x=\frac{8+-\sqrt{24}}{4} \\\\x=\frac{8+-2\sqrt{6}}{4}\\\\x_{1} =2-\frac{\sqrt{6}}{2}\\ \\x_{2} =2+\frac{\sqrt{6}}{2}

7 0
3 years ago
Factorize 5x+6x+3=0<br>​
ruslelena [56]
11x+3=0
11x+3_3=0_3
11x=_3
11x/11=_3/11
x=_3/11
4 0
2 years ago
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