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Serjik [45]
3 years ago
7

Helppp it’s due today!!

Mathematics
1 answer:
Temka [501]3 years ago
6 0

Answer: 15 + c = 64

Step-by-step explanation:

we need to do 15 + 64

C represents her age, which would be 49, if you needed to know

Therefore the equation is 15 + c

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10x - 15 = 12x - 101
GenaCL600 [577]
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Calculate the axis of symmetry in the equation Y = (x+1) (x-4)
algol [13]

Answer:

<em>Every </em><em>parabola </em><em>has </em><em>a </em><em>turning</em><em> point</em><em> </em><em>called </em><em>the </em><em>axis</em><em> of</em><em> symmetry</em><em>,</em><em>so </em><em>for </em><em>this </em><em>question</em><em> </em><em>you </em><em>just</em><em> </em><em>have </em><em>to </em><em>find </em><em>the </em><em>turning</em><em> </em><em>points</em>

<em>x </em><em>turning</em><em> </em><em>point</em><em>=</em><em> </em><em>-b/</em><em>2</em><em>a</em><em>,</em><em>and </em><em>the </em><em>y </em><em>turning</em><em> </em><em>point</em><em> </em><em>is </em><em>gotten</em><em> </em><em>by </em><em>replacing</em><em> </em><em>the </em><em>value </em><em>of </em><em>x </em><em>in </em><em>the </em><em>equation</em>

<em>y=</em><em>(</em><em>x+</em><em>1</em><em>)</em><em>(</em><em>x-4)</em>

<em> </em><em> </em><em>=</em><em>x^</em><em>2</em><em>-</em><em>3</em><em>x</em><em>-</em><em>4</em>

<em>a </em><em>is </em><em>1</em><em>,</em><em>b </em><em>is </em><em>-</em><em>3</em><em> </em><em>and </em><em>c </em><em>is </em><em>-</em><em>4</em>

<em>x=</em><em> </em><em>-</em><em>(</em><em>-</em><em>3</em><em>)</em><em>/</em><em>2</em><em>(</em><em>1</em><em>)</em>

<em> </em><em> </em><em>=</em><em>1</em><em>.</em><em>5</em>

<em>and </em><em>y=</em><em>x^</em><em>2</em><em>-</em><em>3</em><em>x</em><em>-</em><em>4</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>1</em><em>.</em><em>5</em><em>^</em><em>2</em><em>-</em><em>3</em><em>(</em><em>1</em><em>.</em><em>5</em><em>)</em><em>-</em><em>4</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>-</em><em>6</em><em>.</em><em>2</em><em>5</em>

<em>therefore</em><em> </em><em>the </em><em>axis </em><em>of</em><em> symmetry</em><em> is</em><em> (</em><em>1</em><em>.</em><em>5</em><em>,</em><em>-</em><em>6</em><em>.</em><em>2</em><em>5</em><em>)</em>

<em>I </em><em>hope </em><em>this </em><em>helps</em>

6 0
3 years ago
D(x)=(x^2-12x+20)/(3x)
krok68 [10]

Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

<u>Explanation:</u>

d(x) = \frac{x^{2}-12x+20}{3x}

      = \frac{(x-2)(x - 10)}{3x}

Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0  

<u>÷3</u>   <u>÷3 </u>

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

degree of numerator (2) > degree of denominator (1)

so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

  •        <u>(1/3)x - 4    </u>
  •    3x)    x² - 12x + 20
  •             <u>x²        </u>
  •                  -12x
  •                  <u>-12x         </u>
  •                             20 (stop! because there is no "x")

So, slant asymptote is to be drawn at (1/3)x - 4



6 0
3 years ago
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