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hammer [34]
3 years ago
10

A small town experienced an explosive population increase Originally the town had population 170 within 3 years the town's popul

ation increased by 400% what is the town current population
Mathematics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

Step-by-step explanation:

We need to first find out how much 400% of 170 is and then add that increase to the original 170 people.

4(170) = 680 and

680 + 170 = 850 people after 3 years.

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Please solve this thank you!
SSSSS [86.1K]

Answer:

1024x^{40}

Step-by-step explanation:

So the first step is to add like terms since you can simplify the numerator by adding the two values sine they have the same variable and degree.

Add like terms:

[\frac{8x^9}{2x}]^5

Divide by 2x (divide coefficient by 2, subtract coefficient degrees)

[4x^8]^5

Multiply exponents and raise 4 to the power of 5

1024x^{40}

The reason you multiply exponents is because you can think about it like this:

(4 * x * x * x * x * x * x * x * x) (this has one 4 and 8 x's because x is raised to the power of 8. Now if you do that 5 times which is what the exponent is doing you're going to have 40 x's and 8 4's. So it's essentially

(4 * x * x * x * x * x * x * x * x) * (4 * x * x * x * x * x * x * x * x) * (4 * x * x * x * x * x * x * x * x) * (4 * x * x * x * x * x * x * x * x) * (4 * x * x * x * x * x * x * x * x). If you group like terms you'll get (4 * 4 * 4 * 4 * 4) *  (x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x * x). Which simplifies to 4^5 * x ^ (8 * 5) which further simplifies to the answer 1024x^40

6 0
2 years ago
Read 2 more answers
Which graph represents the function?
mario62 [17]
For this function we can find y-intercept.
x=0, y=-2
This graph is on the top, right.
3 0
3 years ago
Someone PLease please help me
makvit [3.9K]
Answer is 1 . Righttttttttttttt
4 0
3 years ago
Help help help if u can answer this I’ll mark u brainliest (6) (the one in the middle)
Hoochie [10]

You just have to plug in each thing:

So f(0):

3(0)^2 - 4 = -4

so its not A

f(-2) and f(2):

3(-2)^2 - 4

3(2)^2 -4

now we don't even have to calculate these, anything to the power of an even number is a positive number so -2^2 and 2^2 both equal 4

B is correct, so you can do the others if you want to check, but if B is true the others shouldn't be true.

4 0
3 years ago
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
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