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Elodia [21]
3 years ago
12

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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7 0
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.

</span>\sf 24, 25, 27, 36, 37, 42, 42, \boxed{\sf 43}, 49, 49, 50, 53, 59, 61, 65

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

\sf 24, 25, 27, \boxed{\sf 36}, 37, 42, 42

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\sf 49, 49, 50, \boxed{\sf 53}, 59, 61, 65

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\sf IQR=Q3-Q1=53-36=\boxed{\sf 17}
3 0
3 years ago
Which in is it.........
ANTONII [103]
I haven't done these problems in a really long time, so there's a chance that my answer might be wrong. I hypothesize that the correct answer is C. 
8 0
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

6 0
3 years ago
Now they know I don't know math
Mumz [18]

Answer:

utzyidfid5dutdgjxjtd75ecb. jsu5sutxjrz74

4 0
3 years ago
Read 2 more answers
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