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sattari [20]
3 years ago
14

There is a quadrilateral MNPQ in which side MN is congruent to side PQ and side NP is parallel to side MQ. The diagonal MP and t

he diagonal NQ intersect each other at point R. If MP = 6x − 5, QR = 3x + 1, and RN = 6, what is QN?
Mathematics
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer: QN = 12

Step-by-step explanation: This quadrilateral is a paralelogram because its 2 opposite sides (NP and MQ) are parallel and the other 2 (MN and PQ) are congruent.

In paralelogram, diagonals bisect each other, which means QR = RN.

If QR = RN:

QR = 6

Then,

QN = QR + RN

QN = 6 + 6

QN = 12

<u>The diagonal QN of quadrilateral MNPQ is </u><u>QN = 12</u><u>.</u>

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> What is the value of x?
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<u>Step-by-step explanation:</u>

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3 years ago
Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7
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Answer:

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Step-by-step explanation:

f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}

slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1

b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}

=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}

8 0
3 years ago
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Likurg_2 [28]

Answer:

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<em>k</em><em>=</em><em>1</em><em>0</em><em>/</em><em>2</em><em>0</em>

<em>k</em><em>=</em><em>1</em><em>/</em><em>2</em>

<em>therefore</em><em> </em><em>relationship</em><em>:</em><em> </em><em>y</em><em>=</em><em>1</em><em>/</em><em>2</em><em>x</em>

<em>subst</em><em> </em><em>x</em><em>=</em><em>1</em><em>5</em><em> </em><em>into</em><em> </em><em>the</em><em> </em><em>relationship</em>

<em> </em><em>y</em><em>=</em><em>1</em><em>/</em><em>2</em><em>(</em><em>1</em><em>5</em><em>)</em>

<em>y</em><em>=</em><em>7</em><em>,</em><em>5</em>

Step by step explanation:

  • Step 1: when they say y varies directly with x they mean<em> y is proportional to x</em>
  • step 2: so y=kx where <em>k is the constant</em>
  • step 3: is to substitute <em>y=10</em> and <em>x=20</em> into the above equation y=kx
  • step 4: you will end up with <em>10=k20</em> then divide both sides by 20 so that <em>k becomes the subject of the formula </em>
  • step 5: your answer from the above will be <em>k=10/20 </em>so the relationship is <em>y is directly proportional to 1/2 x </em>what you did here is that you substituted k for 1/2 in the equation in step 3
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4 0
3 years ago
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