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Vladimir [108]
3 years ago
13

What is the answer? Please Help!

Mathematics
1 answer:
Galina-37 [17]3 years ago
3 0


What is the question you are struggling with I will try and help.


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Ed's hair was 4 centimeters long after his last hair cut, and it grows 11 centimeters every year. How many years will it take un
RSB [31]

Answer:

4

Step-by-step explanation:

48-4=44    44/11=4 therefore the answer is 4.

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Please help Picture can someone help me I don't know how to do this
lorasvet [3.4K]

Answer:

heres the answer

Step-by-step explanation:

just graph (2,3)   (3,2)   (4,3)

on a coordinate grid

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3 years ago
I need to know the opposite of 8 1/4
masya89 [10]

Well Hello!


I believe the answer to your question would be -8 1/4

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3 years ago
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Solve for x -6x + 7 > 1
Elenna [48]

the answer is 6/5

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4 years ago
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Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
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