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3 years ago

What substitution should be used to rewrite 6(x + 5)2 + 5(x + 5) – 4 = 0 as a quadratic equation?

Mathematics

2 answers:

Reika [66]3 years ago

7 0

Answer:

u=(x+5)

Step-by-step explanation:

tiny-mole [99]3 years ago

3 0

Answer:

A: u = (x + 5)

Edg 2020

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3 years ago

What is the interquartile range of the data? 24, 25, 27, 36, 37, 42, 42, 43, 49, 49, 50, 53, 59, 61, 65 41 7 17 10

AysviL [449]

<span>The interquartile range is the difference between the third quartile and the first quartile. First, find the median so you can separate the data in half. The median is the middle number. Arrange the numbers from least to greatest and find the number in the middle.

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Find the medians of these two halves. These will be Quartile 1 and Quartile 3.

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3 years ago

Which in is it.........

ANTONII [103]

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3 years ago

Simplify and answer the boxes.

s2008m [1.1K]

Answer:

$\huge\boxed{\dfrac{x^2+9^2}{x-3y}+\dfrac{6xy}{3y-x}=x-3y}$

Step-by-step explanation:

Domain:

$x-3y\neq0\Rightarrow x\neq3y$

$\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{3y-x}=\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{-(x-3y)}\\\\=\dfrac{x^2+9y^2}{x-3y}-\dfrac{6xy}{x-3y}=\dfrac{x^2+9y^2-6xy}{x-3y}\\\\=\dfrac{x^2-2(x)(3y)+(3y)^2}{3y-x}=\dfrac{(x-3y)^2}{3y-x}\\\\=\dfrac{\bigg[-1(3y-x)\bigg]^2}{3y-x}=\dfrac{(-1)^2(3y-x)^2}{3y-x}\\\\=\dfrac{1(x-3y)(x-3y)}{x-3y}=x-3y$

Used:

The distributive property: a(b + c) = ab + ac

(a - b)² = a² - 2ab + b²

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3 years ago

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Answer:

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3 years ago

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