How to solve this one

Mathematics

2 answers:

Blizzard [7]3 years ago

8 0

y=2-3x

6x+2(2-3x)=4

6x+4-6x=4

6x-6x+4-4=0

0=0

question is mistake

kondor19780726 [428]3 years ago

7 0

Answer:

I think the question has an error

Step-by-step explanation:

As seen in what I've attached, you're supposed to times the top row by 6 and the bottom one by 3 but they both have the same answer so when you go to subtract them from each other it gives you 0=0, I'd suggest to email your teacher

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in thailand the population of town is 48700 men and 41600 women what percentage of the total percentage are men

Finger [1]

Answer:

53.931339977852% (you can round)

Step-by-step explanation:

You add up 48,700 and 41,600 and you would get 90,300 as the overall population. Then you would put it in fraction form if you want to convert it into a percentage. The fraction form would be 48,700/90,300. You could simplify it to 487/903. Then you do 487 x . You would cross multiply and get

___ = ___

903 100

903x = 48700. Then you would divide each side by 903 and get x = 53.931339977852% (you can round)

6 0

2 years ago

Read 2 more answers

Solve the equation for x∈Z <br><br>-x² +8x -14 ≥ 0

kenny6666 [7]

$ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta < 0\ then\ no\ solutions\\if\ \Delta=0\ then\ one\ solution:x=\frac{-b}{2a}\\if\ \Delta > 0\ then\ two\ solutions:x=\frac{-b\pm\sqrt\Delta}{2a}\\================================\\\\-x^2+8x-14\geq0\ \ \ \ |multiply\ both\ sides\ by\ (-1)\ \{change\ \geq\ on\ \leq\}\\\\x^2-8x+14\leq0\\\\a=1;\ b=-8;\ c=14\\\\\Delta=(-8)^2-4\cdot1\cdot14=64-56=8 > 0\\\sqrt\Delta=\sqrt8=\sqrt{4\cdot2}=\sqrt4\cdot\sqrt2=2\sqrt2$

$x_1=\frac{-(-8)-2\sqrt2}{2\cdot1}=\frac{8-2\sqrt2}{2}=\frac{8}{2}-\frac{2\sqrt2}{2}=\boxed{4-\sqrt2}\\\\x_2=\frac{-(-8)+2\sqrt2}{2\cdot1}=\frac{8+2\sqrt2}{2}=\frac{8}{2}+\frac{2\sqrt2}{2}=\boxed{4+\sqrt2}\\\\============================$

$ax^2+bx+c=0\\\\if\ a > 0\ then\ the\ parabola\ o pen\ up\\if\ a < 0\ then\ the\ parabola\ o pen\ down\\========================\\\\a=1 > 0-therefore\ o pen\ up\ (look\ at\ the\ picture)\\\\===============================\\\\Answer:x\in\left$

$Solutions\ in\ \mathbb{Z}:\\\\\sqrt2\approx1.4\\\\threfore:4-\sqrt2\approx4-1.4=2.6\ and\ 4+\sqrt2\approx4+1.4=5.4\\\\look\ at\ the\ second\ picture:x=3\ or\ x=4\ or\ x=5\ (x\in\{3;\ 4;\ 5\})$