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Alex777 [14]
4 years ago
15

Solve I2 x- 2I < 8O [xl-3 <x<5)XIX<-3 or x>5){xl-5 <x<3)​

Mathematics
1 answer:
azamat4 years ago
8 0

Answer: Its b also y u test so ez what grade u in lol?

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LM is 3x, MN is 2x+2. M is the midpoint of LN. What is LM?
Blababa [14]

Answer:

1/2

Step-by-step explanation:

3x=(1/2)(2x+2)

3x=x+1

3x-x=x-x+1

2x=1

x=1/2

7 0
3 years ago
Given the Formula E equals IR what is the formula R
scZoUnD [109]
E = IR
R = E/I

You have to divide I form R
5 0
3 years ago
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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%
andrey2020 [161]

First observe that if a+b>0,

(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}

Let a=0 and b=x. It follows that

a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}

Now let b=1, so a^2+a=4x. Solving for a,

a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2

which means

a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}

Now solve for x.

x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2

(note that we assume 2x-1\ge0)

4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}

(we omit x=0 since 2\cdot0-1=-1\ge0 is not true)

3 0
2 years ago
What is seven less than three times a number.
sukhopar [10]
Let’s use y as a replacement for a number. You would do 3y-7 since you multiply by three and take away seven.
3 0
3 years ago
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What two numbers add to 92 and multiply to 160?
KatRina [158]
None But, 80 x 2 = 160 and 80 + 2 = 82, are you sure you mean 92?

3 0
3 years ago
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