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konstantin123 [22]
3 years ago
14

Yessir help a girl out!

Mathematics
1 answer:
REY [17]3 years ago
4 0

Answer:

3rd option (not regular)

Step-by-step explanation:

all the other options are not true

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What is 4 3/7 as a decimal.
Ymorist [56]
4.428571
I know because I googled it
4 0
2 years ago
Find the vertical, horizontal, and slant asymptotes, if any for y=x^3/ (x-2)^4
VashaNatasha [74]
This has both a horizontal and a vertical asymptote.  There are no slant (oblique) asymptotes cuz the degree of the numerator is not higher than that of the denominator.  If the degree of the numerator is less than the degree of the denominator, which is our case here, then the horizontal asymptote is 0.  But we also have a vertical asymptote, which occurs where the denominator = 0.  We all know that we break every rule known to mankind if we try to divide by 0, so there also a vertical asymptote at x = 2.
7 0
2 years ago
Solve the quadratic equation (2x-3)^2 = 6(3-2x)
LenKa [72]

Answer:

x=\frac{3\sqrt{2}-3 }{2}

x=-\frac{3\sqrt{2}-3 }{2}

Step-by-step explanation:

Open the brackets first:

4x^2+9=18-12x

Simplify:

4x^2+12x-9=0

It is determined that this can't be factored, so rewrite the equation so you can complete the square.

x^2+3x=\frac{9}{4}

Take the square of 1.5 and add it to both sides:

x^2 + 3x+ \frac{9}{4} =\frac{9}4} +\frac{9}{4}

Factor:

(x+\frac{3}{2})^2=\frac{9}{2}

x+\frac{3}{2} = \frac{3\sqrt{2} }{2 }

or x+\frac{3}{2} =-\frac{3\sqrt{2} }{2}

So x=\frac{3\sqrt{2}-3 }{2}

or x=-\frac{3\sqrt{2}-3 }{2}

5 0
2 years ago
One more question plz
oee [108]
1200g
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6 0
2 years ago
HELP ME WITH MATH PLS
Zanzabum

Answer: x = -\frac{10}{3} - \frac{2y}{3}

Step-by-step explanation:

-3x  -2y = 10      Get x alone by add 2y to both sides

      +2y    +2y

-3x = 10  + 2y    Divide both sides by -3

x = -10/3  - 2y/3

4 0
3 years ago
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