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mr Goodwill [35]
2 years ago
5

Quick question. solve for x. show all work

Mathematics
2 answers:
kkurt [141]2 years ago
6 0
75.................................
Ipatiy [6.2K]2 years ago
6 0
75 Because someone said so
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nataly862011 [7]

Step-by-step explanation:

A is the answer to the question

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2 years ago
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What is the word phrase for 5x+2
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Two plus the product of 5 and x
                      or
the product of 5 and x plus 2
                     or
5 times x plus two
5 0
3 years ago
Which side lengths form a right triangle?
vampirchik [111]

Answer:

A and C

Step-by-step explanation:

<h3>A</h3>

=  \sqrt{ {5}^{2} + ( \sqrt{6})^{2}  }  \\  =  \sqrt{25 + 6}  \\  =  \sqrt{31}

<h3>C</h3>

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2 years ago
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In a certain year the price of gasoline rose by 20% during January, fell by 20% during February, rose by 25% in March, and fell
zavuch27 [327]

Answer:

Fell by 25%

Step-by-step explanation:

+20% January

-20% February (20-20 is zero, so we're back to the original price we started at here)

+25% March

?? April

Currently, we are 25% more than what we had, so to get back to normal, the price needs to go down by 25 percent. This will put us back to zero. Put another way...

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8 0
3 years ago
show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this
VLD [36.1K]

Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real

• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

x² - bx - ax - k² = 0 ( in standard form )

x² + x(- a - b) - k² = 0

with a = 1, b = (- a - b), c = -k²

b² - 4ac = (- a - b)² + 4k²

For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R


           

8 0
3 years ago
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