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Stolb23 [73]
3 years ago
10

P=q+2 2p+3q=-26 help

Mathematics
2 answers:
Goshia [24]3 years ago
5 0
P=q+2 making 2p+3q=-26 become:

2(q+2)+3q=-26  perform indicated multiplication on left side

2q+4+3q=-26   combine like terms on left side

5q+4=-26  subtract 4 from both sides

5q=-30  divide both sides by 5

q=-6, and from earlier we know that p=q+2 so

p=-4

Thus (p,q)=(-4,-6)
kompoz [17]3 years ago
3 0
All you have to do is plug and solve...

2(q+2) + 3q = -26
2q + 4 + 3q = -26

Subtract 4 from the left side and move it to the right...

 2q + 4 + 3q = -26
       - 4            - 04
_________________
2q + 3q = -30

Now add 3q and 2q together and you get 5q, next divide...

5q = -30
_______
5         5

Finally, you get q equal to negative 6 (-6)!
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Answer:

Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say \displaystyle \lim_{x\rightarrow a}f(x)=L, if f(x) approaches L as x approaches 'a'.

(a)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\

\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=4\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4\displaystyle \lim_{x\rightarrow 5}x-4(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=4(5)-4(5)\\

\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

(b)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\

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3 0
3 years ago
The number y of deer in a county after x years can be modeled by y=0.75x^2 + 10.5x + 1300. When were there about 2000 deer? Roun
Lemur [1.5K]

Answer:

24 years

Step-by-step explanation:

We are told that the model of number of deers is;

y = 0.75x² + 10.5x + 1300

Where;

y represents number of deers

x represents number of years

Now, we want to find how many years it will take to get to 2000 deers.

Thus;

2000 = 0.75x² + 10.5x + 1300

0.75x² + 10.5x + 1300 - 2000 = 0

0.75x² + 10.5x - 700 = 0

Using quadratic formula, we have;

x = [-b ± √(b² - 4ac)]/2a

x = [-10.5 ± √(10.5²- (4*0.75*-700)]/(2*0.75)

x = [-10.5 ± √(110.25 + 2100)]/(1.5)

x = [-10.5 ± √2210.25]/(1.5)

x ≈ -38 or 24

Number of years can't be negative, so we will take the positive value.

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3 years ago
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stira [4]

I hope this helps you

x=10 f(10)=|10+5|=|15|=15

x=-5 f(-5)=|-5+5|=|0|=0

x=0 f(0)=|0+5|=|5|=5

3 0
3 years ago
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dsp73

Answer:

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Step-by-step explanation:

I took the test hope this helps

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Step-by-step explanation:

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