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3 years ago

Which of the following problems can be solved using the equation below? x+15=2x

Mathematics

1 answer:

pshichka [43]3 years ago

6 0

Answer:

Answer is below

Step-by-step explanation:

Subtract 2x from both sides.

$x+15-2x=2x-2x\\-x+15=0$

Subtract 15 from both sides.

$-x+15-15=0-15\\-x=-15$

Divide both sides by -1.

$\frac{-x}{-1}=\frac{-15}{-1} \\x=15$

So therefore your answer is 15

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What is the slope of a line that is perpendicular to the line whose equation is 2y = 3x - 1?

alexandr1967 [171]

Answer:

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Step-by-step explanation:

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Therefore the slope of the perpendicular line is $-\frac{1}{m}$ .

For our case $y=\frac{3}{2}x-\frac{1}{2}$ therefore the slope of the prependicular is the reciprocal of $\frac{3}{2}$ multiplied by $-1.$

$\therefore\:slope=-\frac{2}{3}.$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Determine the type and number of solutions of 4x^2-5x+1=0.

Arlecino [84]

<span>4x² - 5x + 1 = 0
a = 4; b = - 5, c = 1
</span>Δ = b² - 4.a.c
Δ = (-5)² - 4.4.1
Δ = 25 - 16
Δ = 9
<span>
x = - b </span>± √Δ / 2.a
$x = \frac{-5\± \sqrt{9} }{2*4}$
$x = \frac{-5\±3}{8}$
$x' = \frac{-5-3}{8} = \frac{-8}{8} = -1$
$x' = \frac{-5+3}{8} = \frac{-2}{8} (\div2) = \frac{-1}{4}$

A. Two real solutions (x' = -1 and x'' = -1/4)

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