3 years ago

6a+(_3b)_(_a)_b is =2 and b=_3.

Mathematics

1 answer:

Kruka [31]3 years ago

8 0

Answer:1 2/6

Step-by-step explaination:

6a+(3×2)+2

6a+6+2

6a=8

a=8÷6

a=1 2/6

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Find A and B

mash [69]

Thank you sa points lab you

5 0

2 years ago

Write the equation of a line passing through (-4.3) and perpendicular to -6x + 4y =-12.

sergij07 [2.7K]

Answer:

y=-2/3*x+1/3

Step-by-step explanation:

-6x+4y=-12

put into y=mx+b line equation

4y=6x-12

y=(6/4)x-(12/4)

y=(3/2)x-4

slope is 3/2

to find perpendiular line slope take the negative inverse.

-(2/3) = slope of perpendicular line

use line equation and solve for y-intercept (b)

3=(-2/3)*(-4)+b

3=8/3+b

b= 9/3-8/3 = 1/3 : 3 = 9/3 to allow for subrtaction of fractions.

3 0

3 years ago

H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions?

zubka84 [21]

$H(x)=(x^2)+1\ \ \ \ and\ \ \ \ K(x)=-(x^2)+4. \\\\K(H)=-(H^2)+4=-(x^2+1)^2+4=4-(x^2+1)^2=\\\\.\ \ \ =2^2-(x^2+1)^2=(2-x^2-1)(2+x^2+1)=(1-x^2)(3+x^2)=\\\\.\ \ \ =(1-x)(1+x)(x^2-3\cdot i^2)=(1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)\\\\ K(H)=0\ \ \ \ \Leftrightarrow\ \ \ (1-x)(1+x)(x- \sqrt{3} \cdot i)(x+ \sqrt{3} \cdot i)=0\\\\x=1\ \ \ \ or\ \ \ \ x=-1\ \ \ \ or\ \ \ \ x=\sqrt{3} \cdot i\ \ \ \ or\ \ \ \ x=-\sqrt{3} \cdot i\\\\Ans.\ e.$