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tamaranim1 [39]
4 years ago
13

Ashton says that the zeros of the equation y=6x^2−5x−25 are at x=5/2 and x=−5/3. Which statement correctly describes Ashton's cl

aim? Ashton is incorrect. The quadratic expression has the factors (2x+5) and (3x−5), so the zeros are at x=−5/2 and x=5/3. Ashton is correct. The quadratic expression has the factors (2x+5) and (3x−5), so the zeros are at x=5/2 and x=−5/3. Ashton is incorrect. The quadratic expression has the factors (2x−5) and (3x+5), so the zeros are at x=−5/2 and x=5/3. Ashton is correct. The quadratic expression has the factors (2x−5) and (3x+5), so the zeros are at x=5/2 and x=−5/3.
Mathematics
1 answer:
jeka57 [31]4 years ago
5 0

Answer:

Ashton is correct. The quadratic expression has the factors (2x−5) and (3x+5), so the zeros are at x=5/2 and x=−5/3.

Step-by-step explanation:

The answer itself has the step-by-step explanation.

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E because 8*8=64 so 8/49*8/49 = 64/49
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Is -12.5 greater than -12? will brainlist
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Answer:

No

Step-by-step explanation:

negative numbers work the opposite of how positive numbers work. Think about it like this. -12 is 12 units less than 0 and -12.5 is 12.5 units less than 0 which means that -12.5 is less than -12

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-x-8y+5

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To one-one functions g and h are defined as follows.
Andreyy89

Answer:

  • 9
  • (x-8)/3
  • -1

Step-by-step explanation:

The inverse function for a set of ordered pairs can be found by swapping the x- and y-coordinates in each pair.

  g^{-1}(-1)=9\qquad\text{from the $g(x)$ ordered pair $(9, -1)$}

__

The inverse of a function expressed algebraically can be found by swapping the x- and y-variables and solving for y.

  h^{-1}(x)\qquad\text{is found from }x=h(y)\\\\x=3y+8\\x-8=3y\\\\y=\dfrac{x-8}{3}\\\\\boxed{h^{-1}(x)=\dfrac{x-8}{3}}

A function of its own inverse returns the original value:

  \boxed{\left(h\circ h^{-1}\right)(-1)=-1}

6 0
3 years ago
(cotx+cscx)/(sinx+tanx)
Butoxors [25]

Answer:   \bold{\dfrac{cot(x)}{sin(x)}}

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)

\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}

6 0
3 years ago
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