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Monica [59]
2 years ago
8

Linear or non linear? Y=2x(the power of 2) +X+4

Mathematics
1 answer:
lisabon 2012 [21]2 years ago
3 0

Answer:

It’s linear

Explanation:

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The answer is A

Step-by-step explanation:


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Hey i need help can sb help me please
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5 1/3 + 9 2/3

Step-by-step explanation:

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2 years ago
Solve for x.<br> 7(1x - 3) = 4(x + 5)
alexandr1967 [171]

Answer:

x = 41/3

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS
  • Equality Properties

Step-by-step explanation:

<u>Step 1: Define equation</u>

7(1x - 3) = 4(x + 5)

<u>Step 2: Solve for </u><em><u>x</u></em>

  1. Simplify:                                        7(x - 3) = 4(x + 5)
  2. Distribute:                                     7x - 21 = 4x + 20
  3. Subtract 4x on both sides:          3x - 21 = 20
  4. Add 21 on both sides:                  3x = 41
  5. Divide 3 on both sides:                x = 41/3

<u>Step 3: Check</u>

<em>Plug in x to verify it's a solution.</em>

  1. Substitute:                    7(1(41/3) - 3) = 4(41/3 + 5)
  2. Multiply:                        7(41/3 - 3) = 4(41/3 + 5)
  3. Subtract/Add:               7(32/3) = 4(56/3)
  4. Multiply:                        224/3 = 224/3

Here, we see that 224/3 is indeed equivalent to 224/3. ∴ x = 41/3 is a solution to the equation.

And we have our final answer!

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2 years ago
233.356 round to two significant figures
olchik [2.2K]
The answer is 230 :) 
6 0
3 years ago
Evaluate the cosine if the angle of rotation which contains the point (9, -3) on its terminal side
Liono4ka [1.6K]

so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant

\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{9}}{\stackrel{hypotenuse}{\sqrt{90}}}\implies \stackrel{\textit{rationalizing the denominator}}{\cfrac{9}{\sqrt{90}}\cdot \cfrac{\sqrt{90}}{\sqrt{90}}\implies \cfrac{9\sqrt{90}}{90}}\implies \cfrac{\sqrt{90}}{10}\implies \cfrac{3\sqrt{10}}{10}

6 0
3 years ago
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