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Alona [7]
3 years ago
12

Use the quadratic formula to find solutions to the quadratic equation given below. 2x2+x-1=0

Mathematics
1 answer:
stiv31 [10]3 years ago
8 0

<u>Answer</u>:

\boxed{x = \frac{1}{2}} \text{and} \boxed{x = - 1}

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

The quadratic formula is: $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

So plugging in all the values, we are left with \frac{-1 \pm 3}{4}.

Then, we can solve for both values of x, which are x = 1/2 and x = -1

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Find the absolute extrema of the function over the region R. (In each case, R contains the boundaries.) Use a computer algebra s
algol [13]

<em>f(x, y)</em> = <em>x</em> ² - 4<em>xy</em> + 5

has critical points where both partial derivatives vanish:

∂<em>f</em>/∂<em>x</em> = 2<em>x</em> - 4<em>y</em> = 0   ==>   <em>x</em> = 2<em>y</em>

∂<em>f</em>/∂<em>y</em> = -4<em>x</em> = 0   ==>   <em>x</em> = 0   ==>   <em>y</em> = 0

The origin does not lie in the region <em>R</em>, so we can ignore this point.

Now check the boundaries:

• <em>x</em> = 1   ==>   <em>f</em> (1, <em>y</em>) = 6 - 4<em>y</em>

Then

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 6 when <em>y</em> = 0

max{<em>f</em> (1, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -2 when <em>y</em> = 2

• <em>x</em> = 4   ==>   <em>f</em> (4, <em>y</em>) = 12 - 16<em>y</em>

Then

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = 12 when <em>y</em> = 0

max{<em>f</em> (4, <em>y</em>) | 0 ≤ <em>y</em> ≤ 2} = -4 when <em>y</em> = 2

• <em>y</em> = 0   ==>   <em>f</em> (<em>x</em>, 0) = <em>x</em> ² + 5

Then

max{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 21 when <em>x</em> = 4

min{<em>f</em> (<em>x</em>, 0) | 1 ≤ <em>x</em> ≤ 4} = 6 when <em>x</em> = 1

• <em>y</em> = 2   ==>   <em>f</em> (<em>x</em>, 2) = <em>x</em> ² - 8<em>x</em> + 5 = (<em>x</em> - 4)² - 11

Then

max{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -2 when <em>x</em> = 1

min{<em>f</em> (<em>x</em>, 2) | 1 ≤ <em>x</em> ≤ 4} = -11 when <em>x</em> = 4

So to summarize, we found

max{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = 21 at (<em>x</em>, <em>y</em>) = (4, 0)

min{<em>f(x, y)</em> | 1 ≤ <em>x</em> ≤ 4, 0 ≤ <em>y</em> ≤ 2} = -11 at (<em>x</em>, <em>y</em>) = (4, 2)

5 0
3 years ago
What is the leading coefficient of the polynomial when written in standard<br> form?<br> 872 +9+ 5x
lianna [129]

Answer:

5

Step-by-step explanation:

The coefficient is known as "the number in front of the letter,' so since 5 is in front of the x, 5 would be the leading coefficient, especially since the outcome would be 5x+881.

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3 years ago
Yoo i need help !!! its math and i havent learned it yet
BlackZzzverrR [31]

Answer:

{f}^{ - 1} (x) = 9x + 18

Step-by-step explanation:

f(x) =  \frac{1}{9}x - 2

y = \frac{1}{9} x - 2

x =  \frac{1}{9}y - 2

x + 2 =  \frac{1}{9}y

9x + 18 = y

{f}^{ - 1} (x) = 9x + 18

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3 years ago
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If d equals 3, what is the value of 6d-2?
umka2103 [35]

Answer:

16

Step-by-step explanation:

6(3)-2

16

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3 years ago
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