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Elan Coil [88]
2 years ago
10

(Will give Brainliest) Explain what the graph of the function represents. Be sure to use complete sentences.

Mathematics
1 answer:
Irina-Kira [14]2 years ago
4 0

Answer: f(-735) = -\frac{2}{3}(-735) + 490 = 980

Step-by-step explanation:

f(x) = -\frac{2}{3} x + 490 =0  Subtract 490 from both sides

        \frac{3}{2}* -\frac{2}{3} x = -490 *\frac{3}{2}   Then flip -\frac{2}{3} multiply both sides by \frac{3}{2} to cancel out -\frac{2}{3}

      x= -735  Thus giving you -735 as your x.

Then you plug it back into the equation to solve.  

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3 years ago
Evaluate 7y - 2<br> when y = 9
kotykmax [81]

Answer:

The answer is 61.

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
If a quartic equation has a real root, what do you know about the other roots of the equation? Explain.
Alex17521 [72]

Answer:

A) Real Roots

B) Imaginary Roots

C) Equal Roots

D) Unequal Roots

E) Rational Roots

F) Irrational Roots

Step-by-step explanation:

The equation with one variable , in which the highest power of variable is two , is know as QUADRATIC EQUATION

ex - 3x² + 4x + 7 = 0

Every quadratic equation gives to values of unknown variables and these values is called roots of equation .

The quadratic equation have three roots :

A) Real Roots

B) Imaginary Roots

C) Equal Roots

D) Unequal Roots

E) Rational Roots

F) Irrational Roots

The nature of Roots depends entirely on the value of it Discriminant

If D = 0 ,  Roots are real and equal

If D > 0  , roots are real and unequal

If D <  0 ,roots are imaginary

where D = b² - 4ac for any quadratic equation ax² + bx + c = 0

Answer

6 0
3 years ago
Plz with steps .. it's very hard can anyone plz
liubo4ka [24]

Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}  =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\

7 0
3 years ago
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