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Luba_88 [7]
4 years ago
9

Simplify. 7/11 -3/11

Mathematics
1 answer:
allochka39001 [22]4 years ago
3 0

Answer:

4/11

Step-by-step explanation:

7/11-3/11

=>(7-3)/11

=>4/11

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Will give brainliest please help ASAP
borishaifa [10]

Answer:

D is the answer. When you graph the function you get the last graph



8 0
3 years ago
Read 2 more answers
33 = -u/4<br><br> please help me i really confused......
nordsb [41]

Answer:

u = -132

Step-by-step explanation:

33=-u/4

33*4=-u

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132/-1=u

-132=u

u = -132

Check:

33= -1(-132)/4

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Hope this helps. Have a nice day.

6 0
3 years ago
Let f(x)=16x^2-x. Find f(x+1)-f(x)
olasank [31]

f(x+1) = 16(x+1)² - (x+1) = 16(x²+2x+1) - x - 1

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so

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8 0
3 years ago
169= (24/2)^2 +(d/2)^2
OlgaM077 [116]

Answer: d = ±10

<u>Step-by-step explanation:</u>

169 = (\frac{24}{2})^{2} + (\frac{d}{2})^{2}

169 = (12)² + \frac{d^{2}}{4}

169 = 144  + \frac{d^{2}}{4}

25 =            \frac{d^{2}}{4}

100 = d²

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6 0
3 years ago
The roots of 2x^{2} + 3x = 4 are α and β. Find the simplest quadratic equation which has roots <img src="https://tex.z-dn.net/?f
elena-s [515]
2x^2+3x=4\implies2x^2+3x-4=0\implies x=\dfrac{-3\pm\sqrt{41}}4

Let \alpha be the root with the positive square root and \beta the root with the negative square root. Then

\dfrac1\alpha=\dfrac4{-3+\sqrt{41}}\quad\text{and}\quad\dfrac1\beta=\dfrac4{-3-\sqrt{41}}

The simplest quadratic with these roots can be written as

\left(x-\dfrac1\alpha\right)\left(x-\dfrac1\beta\right)=x^2-\left(\dfrac1\alpha+\dfrac1\beta\right)x+\dfrac1{\alpha\beta}=x^2-\dfrac34x-\dfrac12
4 0
3 years ago
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