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grigory [225]
3 years ago
9

Write the numerical expression 34 in expanded form.

Mathematics
1 answer:
garik1379 [7]3 years ago
5 0

Answer:

not sure what type you mean but here are options

Step-by-step explanation:

30+4=34

thirty-four

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Then, we simplify this.
-31x-32=0
Then, isolate x
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Y<1-6x


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4 0
3 years ago
Consider the system of equations given in slope-intercept form. y = −13x + 17, y = 5x - 23 The solution seems to be about (8, 14
Nitella [24]

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The exact solution for this equation would be (2.222, -11.889)

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3 years ago
Solve −12 − 2x &lt; −3x + 2 &lt; −2(x − 7). 14 over 5 &lt; x &lt; negative 12 over 5 −12 &lt; x &lt; 14 14 &lt; x &lt; −12 12 ov
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-12-2x < -3x+2 < -2(x-7)\\\\-12-2x < -3x+2 < -2x+14\ \ \ \ |+2x\\\\-12 < -x+2 < 14\ \ \ \ |-2\\\\-14 < -x < 12\ \ \ \ |\text{change the signs}\\\\14 > x > -12

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3 years ago
Solve by quadratic equation​
Ymorist [56]
<h2>Question :</h2>

  • \tt \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}

<h2>Answer :</h2>

  • \large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}

<h2>Explanation :</h2>

\tt : \implies \dfrac{x+2}{x-2} + \dfrac{x-2}{x+2} = \dfrac{5}{6}

\tt : \implies \dfrac{(x+2)(x+2) + (x-2)(x-2)}{(x-2)(x+2)} = \dfrac{5}{6}

\tt : \implies \dfrac{(x+2)^{2} + (x-2)^{2}}{(x-2)(x+2)} = \dfrac{5}{6}

<u>Now, we know that</u> :

  • \large \underline{\boxed{\bf{(a+b)^{2} = a^{2} + b^{2}+ 2ab}}}
  • \large \underline{\boxed{\bf{(a-b)^{2} = a^{2} + b^{2} - 2ab}}}
  • \large \underline{\boxed{\bf{(a+b)(a-b) = a^{2} - b^{2}}}}

\tt : \implies \dfrac{x^{2}+2^{2}+ 2 \times x \times 2 + x^{2}+2^{2} - 2 \times x \times 2 }{x^{2}-2^{2}} = \dfrac{5}{6}

\tt : \implies \dfrac{x^{2}+ 4 + \cancel{4x} + x^{2}+ 4 - \cancel{4x}}{x^{2}-4} = \dfrac{5}{6}

\tt : \implies \dfrac{x^{2} + x^{2} + 4 + 4}{x^{2}-4} = \dfrac{5}{6}

\tt : \implies \dfrac{2x^{2} + 8}{x^{2}-4} = \dfrac{5}{6}

<u>By cross multiply</u> :

\tt : \implies (2x^{2} + 8)6= 5(x^{2}-4)

\tt : \implies 12x^{2} + 48 = 5x^{2}-20

\tt : \implies 12x^{2} + 48 - 5x^{2} + 20 = 0

\tt : \implies 7x^{2} + 68 = 0

\tt : \implies 7x^{2} + 0x + 68 = 0

<u>Now, by comparing with ax² + bx + c = 0, we have</u> :

  • a = 7
  • b = 0
  • c = 68

<u>By using quadratic formula</u> :

\large \underline{\boxed{\bf{x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}}}

\tt : \implies x = \dfrac{-(0) \pm \sqrt{(0)^{2} - 4(7)(68)}}{2(7)}

\tt : \implies x = \dfrac{0 \pm \sqrt{0 - 1904}}{14}

\tt : \implies x = \dfrac{\pm \sqrt{- 1904}}{14}

\tt : \implies x = \dfrac{\pm \sqrt{2\times 2\times 2\times 2\times 7\times 17}}{14}

\tt : \implies x = \dfrac{\pm \cancel{2} \times 2\sqrt{7\times 17}}{\cancel{14}}

\tt : \implies x = \dfrac{\pm2\sqrt{119}}{7}

\large \underline{\boxed{\bf{x = \dfrac{\pm 2\sqrt{119}}{7}}}}

Hence value of \bf x =\dfrac{\pm 2\sqrt{119}}{7}

5 0
3 years ago
Read 2 more answers
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