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Crank
3 years ago
14

Find the x- and y-intercepts of the equation below. Write your answers as ordered pairs.

Mathematics
1 answer:
Airida [17]3 years ago
5 0

Answer:

x intercept: none

y intercept=: (0,-12)

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Which division problem is equivalent to 2/5
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Cut 5 equal sections from a 2 foot long ribbond
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What is thesum of the two polynomials ? (-x^2+9)+(-3x^2-11x+4)
True [87]
-x^2 + 9 - 3x^2 - 11x + 4
-4x^2 - 11x + 13
3 0
3 years ago
-15.6 = -3.9r<br> please include an explanation of how you got your answer :) thanks
Mumz [18]

Here is the equation:

-15.6 = -3.9r

Since both decimals are negative, the answer will be positive.

  • Negative times Negative = Positive

Divide both sides by -3.9 to leave the variable alone:

-15.6 \div -3.9  = -3.9r \div -3.9 \\ 4 = r

r is equal to 4 → r = 4

5 0
3 years ago
How many solutions does this linear system have?
alexira [117]
The answer here is no solution
5 0
2 years ago
A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. A
Naddik [55]

Answer:

Initial bacterias = 6006000

Altought I believe is safe to assume that the values were 192,000 and 384,000 instead of 192,192,000 and 384,384,000, in that case the initial bacterias is 6000

Step-by-step explanation:

A exponential growth follows this formula:

Bacterias  = C*rⁿ

C the initial amount

r the growth rate

n the number of time intervals

Bacterias (55 hours) = 192,192,000

Bacterias (66 hours) = 384,384,000

Bacterias(55hours)=C*r^{{\frac{55-t}{t}}} \\Bacterias (66hours) = C*r^{\frac{66-t}{t}}}

If you divide both you can get the growth rate:

\frac{Bacterias (66hours)}{Bacterias(55hours)}=\frac{C*r^{\frac{66-t}{t}}}{C*r^{{\frac{55-t}{t}}}} \\\frac{384,384,000}{192,192,000} =r^{\frac{66-t}{t} -\frac{55-t}{t} } \\2 =r^{\frac{11}{t}}

So with that r = 2 and each time interval correspond to 11 years

Then replacing in one you can get the initial amount of C

Bacterias (55hours)=C*2^{\frac{55-11}{11} } 192,192,000 = C*32\\C= 6006000

7 0
3 years ago
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