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3 years ago

13

Could someone help me solve this?

1 answer:

Andrew [12]3 years ago

5 0

G has 2 local minimums, you can tell because there are 2 dips in the line before it goes of to infinity.
Both questions have the same answer so the answer to both would be
Local Min: (-2,-3), (3,0)

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What percent of 3,000 is 18

neonofarm [45]

Answer: 0.6%

Step-by-step explanation:

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3 years ago

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Lisa is making activity baskets to donate to charity. she has 12 coloring books, 28 markers, and 36 Crayons. what is the greates

ahrayia [7]

4 is the highest number that goes into 12,28,and 36

12/4 = 3
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so she can make 4 baskets....each containing 3 coloring books, 7 markers, and 9 crayons

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4 years ago

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Asap - i need help like asap so if someone knows can you help me??

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3 years ago

Tim earns $31 after 2 hours, $46.5 after three hours,

olganol [36]

Answer:

Equation: y = 15.5x

Step-by-step explanation:

Given that:

Time earns $31 for 2 hours

$46.5 for 3 hours

$62 for 4 hours

We will find unit rate first.

Unit rate = $\frac{31}{2}$ = $15.5 per hour

Now,

Let,

x be the number of hours

y be the total money earned by Tim

Total earning = Per hour rate * Number of hours

y = 15.5x

Hence,

Equation: y = 15.5x

4 0

3 years ago

The population of a town doubled approximately every 5 years during the first several decades after it was founded. If the origi

zzz [600]

Answer:

624 people

(I put two ways to look at the problem.)

Step-by-step explanation:

What is describe here is an exponential function of the form:

$P=P_0 e^{kt}$

$t$ is the number of years after 1761.

$P_0$ is the initial population.

So t=0 represents year 1761.

t=1 represents year 1762

t=2 represents year 1763

....

t=20 represents year 1781.

So we have the doubling time is 5 years. This means the population will be twice what it was in 5 years. Let's plug this into:

$P=P_0e^{kt}$

$2P_0=P_0e^{k\cdot 5}$

Divide both sides by $P_0$ :

$2=e^{5k}$

Convert to logarithm form:

$5k=\ln(2)$

Multiply both sides by 1/5:

$k=\frac{1}{5}\ln(2)$

$k=\ln(2^{\frac{1}{5}})$ By power rule.

So in the next sentence they actually give us the initial population and we just found k so this is our function for P:

$P=39e^{\ln(2^{\frac{1}{5}})t}$

So now we plug in 20 to find how many residents there were in 1761:

$P=39e^{\ln(2^{\frac{1}{5}})(20)}$

This is surely going to the calculator:

$P=624$

Now if you don't like that, let's try this:

Year 0 we have 39 people.

Year 5 we have 39(2)=78 people.

Year 10 we have 78(2)=156 people.

Year 15 we have 156(2)=312 people.

Year 20 we have 312(2)=624 people.

7 0

3 years ago