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rewona [7]
3 years ago
12

The Problem is= 17/15 = 18/x + 1

Mathematics
1 answer:
grin007 [14]3 years ago
8 0

Answer:

1/135

Step-by-step explanation:

subtract the 1 from 17/15

you get 2/15 and then you multiply that by 1/18

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The greatest possible side length is 24 square piece
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3 years ago
(NEED HELP WILL MARK BRAINLIEST)
Basile [38]

Answer:

All of the above

Step-by-step explanation:

Also, this is a history, not a maths problem

6 0
3 years ago
What is the discriminant of the quadratic equation 0 = –x2 4x – 2? a.–4 b.8 c.12 d.24
MrMuchimi
X^2 - 4x + 2
d = b^2 - 4ac, where a = 1, b = -4 and c = 2
d = (-4)^2 - 4 x 1 x 2
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4 0
3 years ago
Your help is deeply is appreciated.
MrRa [10]

Answer:

x = 3

Step-by-step explanation:

\\4x^2-2x=30\\\\x=3\\4(3)^2-2(3)=30\\4(9) - 6 =30\\36 - 6 = 30\\30 = 30 \\\\x=4\\4(4)^2-2(4)=30\\4(16) - 8 =30\\64 - 8 = 30\\56\neq 30

56  is not equal to 30

6 0
3 years ago
Read 2 more answers
64, –48, 36, –27, ...<br><br> Which formula can be used to describe the sequence?
nordsb [41]

Answer:

\boxed{a_n \:  =  \: 64 \:  \times  \: ( -  \frac{3}{4} ) ^{n \:  -  \: 1} }

Step-by-step explanation:

  • We first compute the ratio of this geometric sequence.

r \:  =  \:  \frac{ - 48}{64}  \\  \\ r   \:  =  \:  \frac{36}{ - 48}  \\  \\  r \:  =  \:  \frac{ - 27}{36}

  • We simplify the fractions:

r \:  =  \:   -  \frac{3 }{4}   \\  \\ r   \:  =  \:   -  \frac{3 }{4}  \\  \\  r \:  =  \:    -  \frac{3 }{4}

  • We deduce that it is the common ratio because it is the same between each pair.

r \:  =  \:  -  \frac{3 }{4}

  • We use the first term and the common ratio to describe the equation:

a_1 \:  =  \: 64; \: r \:  =  \:  -  \frac{3 }{4}

<h3>We apply the data in this formula:</h3>

\boxed{a_n \:  =  \: a_1 \:   \times  \:  {r}^{ n \:  -  \: 1} }

_______________________

<h3>We apply:</h3>

\boxed {\bold{a_n \:  =  \: 64 \:   \times  \:  {( -  \frac{3}{4} )}^{ n \:  -  \: 1} }}

<u>Data</u>: The unknown "n" is the term you want

<h3><em><u>MissSpanish</u></em></h3>
4 0
2 years ago
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