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AveGali [126]
3 years ago
12

Match each equation to its equivalent equation in slope-intercept form.

Mathematics
1 answer:
Bumek [7]3 years ago
6 0
1.) y+6=3(x+2)    C
     y+6=3x+6
     y=3x+6-6
     y=3x+6
2.) y=1/2(x+8)-2    B
     y=1/2x+4-2
     y=1/2x+2
3.) y+1=1(x-3)    E
     y+1=x-3
     y=x-3-1
     y=x-4
4.) -4x+y=-2    A
     y=-4x-2
5.) 2x-4y=-4     F
     -4y=-2x-4
     y=-2/-4x-4/-4
     y=2/4x+4/4
     y=1/2x+1
6.) 2x+4y=8     D
     4y=-2x+8
     y=-2/4x+8/4
     y=-1/2x+4/2
     y=-1/2x+2
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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 1.83 repeating as a fraction?
maks197457 [2]
The answer to this question would be: 1 5/6

To change a decimal number into fraction, you need to divide the number on the right of decimal point with 1. In this case, the number is 0.83.
This number is hard since .83 doesn't have many factors. To find the answer you can try to multiply the decimal with some number until it close to 1(no decimal left)
0.83* 2= 1.66
0.83* 3= 2.49 ---> close to half, if you find this number, you can try to double it
0.83* 4= 3.32
0.83* 5= 4.15
0.83* 6= 4.98---> close to 1, that means there is high probability that the number can be divided by 6 
0.83 would be 4.98/6, but if we assume that the number is 0.8333...... then 0.83 would be 5/6. So, 1.83 would be 1 5/6
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Aneli [31]

Answer:

Deductive reasoning

Step-by-step explanation:

Here, we want to know the kind of reasoning the statement is.

In this question, the kind of reasoning is deductive

In the deductive reasoning type, we go from general cases to specific cases

It is just like saying the reason why there will be hurricanes in the tropics this fall is because there is an hurricane every fall

So we have gone from a general case to give a specific case

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3 years ago
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