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Akimi4 [234]
3 years ago
9

How do you factor this? x^{2} + 5 x- 9 = ?

Mathematics
1 answer:
Oksanka [162]3 years ago
6 0
x^2+5x-9\\\\a=1;\ b=5;\ c=-9\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot1\cdot(-9)=25+36=61 > 0\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{61}\\\\x_1=\dfrac{-5-\sqrt{61}}{2\cdot1}=\dfrac{-5-\sqrt{61}}{2}\\\\x_2=\dfrac{-5+\sqrt{61}}{2\cdot1}=\dfrac{-5+\sqrt{61}}{2}\\\\x^2+5x-9=\left(x-\dfrac{-5-\sqrt{61}}{2}\right)\left(x-\dfrac{-5+\sqrt{61}}{2}\right)
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Hi, am I supposed to find the stationary points of the f(x) and g(x) then use the distance formula to solve for the final answer
nikdorinn [45]

9514 1404 393

Answer:

  maximum difference is 38 at x = -3

Step-by-step explanation:

This is nicely solved by a graphing calculator, which can plot the difference between the functions. The attached shows the maximum difference on the given interval is 38 at x = -3.

__

Ordinarily, the distance between curves is measured vertically. Here that means you're interested in finding the stationary points of the difference between the functions, along with that difference at the ends of the interval. The maximum difference magnitude is what you're interested in.

  h(x) = g(x) -f(x) = (2x³ +5x² -15x) -(x³ +3x² -2) = x³ +2x² -15x +2

Then the derivative is ...

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This has zeros (stationary points) at x = -3 and x = 5/3. The values of h(x) of concern are those at x=-5, -3, 5/3, 3. These are shown in the attached table.

The maximum difference between f(x) and g(x) is 38 at x = -3.

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The number is <u>20</u>.

Step-by-step explanation:

In this question, it is said that 35% of a number is equal to 7, then what is that number?

So, let the number be <em>x</em>.

Now, according to the question :

\begin{gathered} \begin{array}{l} \quad{\dashrightarrow{\sf{35 \% \: of \: x = 7}}}\\\\\quad{\dashrightarrow{\sf{\dfrac{35}{100} \: of \: x = 7}}}\\\\\quad{\dashrightarrow{\sf{\dfrac{35}{100}  \times x = 7}}}\\\\\quad{\dashrightarrow{\sf{\dfrac{35 \times x}{100} = 7}}}\\\\\quad{\dashrightarrow{\sf{\dfrac{35x}{100} = 7}}}\\\\\quad{\dashrightarrow{\sf{x = 7 \times \dfrac{100}{35}}}}\\\\\quad{\dashrightarrow{\sf{x =\dfrac{700}{35}}}}\\\\ \quad{\dashrightarrow{\sf{x = \cancel{\dfrac{700}{35}}}}}\\\\ \quad{\dashrightarrow{\sf{x =20}}}\\\\ \quad{\star{\underline{\boxed{\tt{\red{x =20}}}}}} \end{array}\end{gathered}

Hence, the number is 20.

\rule{300}{2.5}

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