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Arisa [49]
2 years ago
13

Solve using quadratic formula 6x^2-x=2

Mathematics
1 answer:
Nadya [2.5K]2 years ago
7 0

6x^2-x=2\implies 6x^2-1x-2=0 \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{6}x^2\stackrel{\stackrel{b}{\downarrow }}{-1}x\stackrel{\stackrel{c}{\downarrow }}{-2}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}

x = \cfrac{ -(-1) \pm \sqrt { (-1)^2 -4(6)(-2)}}{2(6)}\implies x = \cfrac{1\pm\sqrt{1+48}}{12} \\\\\\ x = \cfrac{1\pm\sqrt{49}}{12}\implies x = \cfrac{1\pm 7}{12}\implies x = \begin{cases} \frac{8}{12}\to &\frac{2}{3}\\[1em] -\frac{6}{12}\to &-\frac{1}{2} \end{cases}

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Find the input (x) of the function y=5x-3 if the output (y) is 32
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Answer:

x = 7

Step-by-step explanation:

Given

y = 5x - 3 ← equate 5x - 3 to 32

5x - 3 = 32 ( add 3 to both sides )

5x = 35 ( divide both sides by 5 )

x = 7 ← input

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3 years ago
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Step-by-step explanation:

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The measure of the supplement of an angle is seven times as large as the measure of its complement find the measure of the angle
8090 [49]

Answer: original angle = 75°

Its supplement 105°

Its complement = 15°

Step-by-step explanation:

We know that,

  • Sum of supplementary angles is 180 degrees.
  • Sum of complementary angles is 90 degrees.

Let x be the original angle , then its supplement = 180° - x

its complement = 90° - x

As per given,

180° - x = 7(90° - x)

⇒ 180° - x = 630° - 7x

⇒  7x- x = 630° -180°

⇒  6 x = 450°

⇒  x = 75°  [divide both sides by 6]

So, original angle = 75°

Its supplement =  180° - 75° = 105°

Its complement = 90° -75° = 15°

Hence,  original angle = 75°

Its supplement 105°

Its complement = 15°

4 0
3 years ago
Simplify square root of (1-cos)(1+cos)/cos^2
Pavel [41]
I hope this helps you



(1-cosx)(1+cosx)=1+cosx-cosx-cos²x=1-cos²x=sin²x


sin²x
____
cos²x


tg²x
3 0
2 years ago
A = {a, b, c, d}
Helen [10]

Answer:

Johnny is wrong.

A \cup B = \left\{a,b,c,d,e,f,g,h\right\}

Step-by-step explanation:

Johnny is wrong.

A better definition would be: A \cup B is the set of all elements that belong to at least one A or B.

So, the elements that belong to both A and B, like c and d in this exercise, also belong to A \cup B.

So:

A \cup B=\left\{a,b,c,d,e,f,g,h\right\}

4 0
2 years ago
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