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Pachacha [2.7K]
3 years ago
7

I'm horrible at this please help:((((​

Mathematics
1 answer:
wel3 years ago
8 0

Step-by-step explanation:

You can insert different values of x to find y

For example:

x = 4

Plug in x into the equation

y = -1/2(4) + 5

y = -2 + 5

y = 3

So your point will be (4, 3)

If you do 1 more like this, you can draw a line between them. Or you can use Desmos to graph the line.

Hope this helped you!!

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Answer:

The answer is A. 12%

Step-by-step explanation:

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3 years ago
Solve for the value of x:-<br> <img src="https://tex.z-dn.net/?f=4x-5%20%3D%203" id="TexFormula1" title="4x-5 = 3" alt="4x-5 = 3
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4x-5 = 3

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Use the digits 3,7,1 and 5 to write number that round 3500
devlian [24]
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3 years ago
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

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The rates are not equivalent; 24/6 is less than the rate 36/8.

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