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alexandr402 [8]
3 years ago
9

I) Domain, ii) x-intercept, iii) y-intercept, iv) vertical asymptote, and v) horizontal asymptote of

Mathematics
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

i)

Step-by-step explanation:

i) you can't divide by 0 and when you put x=5 into the bottom part 4x-20 = 4(5)-20 = 20-20 = 0 so x cannot be 5. So the domain is x > 5 and x < 5 OR in other words, when x is not 5.

ii) substitute y=0 into the equation and solve the quadratic on top

iii) substitute x=0 into the equation and write down the value

horizontal asymptote at x=5 because we found out that x cannot be 5 in part i)

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(-3/4)/=-3/4•4/3=-12/12=-1
Vaselesa [24]

Answer: (-3/4)/=-3/4•4/3=-12/12=-1 = = 0

Step-by-step explanation:

Unary minus: -3

4

Multiple: the result of step No. 3 * 4

3

= -3

4

* 4

3

= -3 · 4

4 · 3

= -12

12

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1 · 12

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Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(-12, 12) = 12.

In words - minus three quarter multiplied by four thirds = minus one.

Compare: (the result of step No. 4) = (-1)

Unary minus: -12

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Compare: (the result of step No. 6) = (-1)

Compare: (-1)

7 0
3 years ago
A complete table with values for x or y that make this equation true:3x+y=15
Andreyy89

Answer:


Step-by-step explanation:

The given equation is: 3x+y=15,

Now, putting x=2 in above equation,

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y=15-6=9

Now, put y=3 in the given equation,

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x=4

Putting x=6,

3(6)+y=15

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Putting x=0,

0+y=15

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Putting x=3,

3(3)+y=15

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Putting y=0,

3x+0=15

x=5

Putting y=8,

3x+8=15

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4 0
3 years ago
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Brrunno [24]
21     70
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4 years ago
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torisob [31]

Answer:

6

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\frac{DC}{4}=\frac{12}{8} \\ \\ DC=6

3 0
1 year ago
The sum of two consecutive positive odd integers is two hundred sixty-four. find the two integers.
Daniel [21]

Any integer (not a fraction) that cannot be divided exactly by 2 is an odd integer.

So, we have ( 2k + 1 ) + ( 2k + 3 ) = 264, where 2k + 1 and 2k + 3 are the two consecutive positive odd integers ( k > 0 );

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3 years ago
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