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Step2247 [10]
3 years ago
13

(3x−2

formula">+(2x−4)^{2} =3
please solve
Mathematics
2 answers:
bazaltina [42]3 years ago
8 0
Infinite solutions (extra characters)
Tpy6a [65]3 years ago
7 0
Question: (3x-2)² + (2x-4)² = 3

Answer:

When squaring things, we are just multiplying it by itself (a² = a x a). When squaring brackets, it’s the same rule:

Therefore:

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

And using this skill we get:

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

Using FOIL:

9x²-6x-6x+4 + 4x²-8x-8x+16 = 3

Simplify:

13x²-28x+20 = 3

Take -20 off from both sides:

13x²-28x = -17

There’s your answer!

I hope this made sense, if you have any questions or you think I made a mistake somewhere please let me know!

:)
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jonny [76]
-0.25 would be -25/100 which can be reduced to -1/4
3 0
3 years ago
when you sign up for video games are you get 400 points you earn 50 more points for each hour that you play how many hours do yo
aleksandr82 [10.1K]
This is a linear equation
y=mx+b
y=50x+400 (when x equals the number of hours played)

1000=50x+400
*first, isolate the variable (x)
-50x=-1000+400
*divide by -50 to get x by itself
x=20+8

The answer is x=28
4 0
3 years ago
Read 2 more answers
Which property will Sarah use to solve 6x = 42?
grin007 [14]

Answer:

multiplication property of equality

Step-by-step explanation:

6 0
3 years ago
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What are the solutions to the equation
frosja888 [35]

Answer:

C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

Step-by-step explanation:

You have the quadratic function 2x^2-x+1=0 to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions ax^2+bx+c=0 with a\neq 0 the Bhaskara's Formula is:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}

x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}

It usually has two solutions.

Then we have  2x^2-x+1=0  where a=2, b=-1 and c=1. Applying the formula:

x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}

Observation: \sqrt{-1}=i

x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i

And,

x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i

Then the correct answer is option C.

x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i and x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i

3 0
3 years ago
Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!
Kitty [74]

Answer & Step-by-step explanation:

In the problem, we are given an equation. h(t) = -20 + 11t

In the equation, t represents an unknown value. So, if we are given a number that is in the replacement of t, then we can plug that number in to where t is at.

t = 11

h(11) = -20 + 11(11)

h(11) = -20 + 121

h(11) = 101

So, h(11) is equal to 101

3 0
3 years ago
Read 2 more answers
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