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xxMikexx [17]
3 years ago
11

How much normally is retaking a high school class?

Mathematics
1 answer:
leonid [27]3 years ago
5 0

Answer:

What do you mean?

Step-by-step explanation:

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What are the coordinates of the point obtained if the point A=(3,7) is rotated 180 degrees about the origin?
Yuliya22 [10]

Answer:

B. (-3, -7)

Step-by-step explanation:

Anytime you rotate a number 180° about the origin, the numbers in the coordinates become their opposites.

-3 is the opposite of 3 and -7 is the opposite of 7.

Therefore, your new coordinates are (-3, -7)

(x,y) becomes (-x,-y)

8 0
3 years ago
Simplify 3(2f+6)+7<br><br><br><br><br><br><br><br>thanks
Naya [18.7K]

{Expand} 3(2f+6): 6f+18

6f+18+7

18+7=25

= 6f+25

4 0
3 years ago
(3x + 5)(3x – 6)<br> 18x – 6<br> 9x² – 3x - 30<br> 9x? 334 – 30<br> 6x - 1
8090 [49]

Answer:

v

Step-by-stedcvp explanation: welcome to family fued

fvdvfdvvdvvvvvvvvvvv

6 0
3 years ago
A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha
8_murik_8 [283]
To model this situation, we are going to use the decay formula: A=Pe^{rt}
where 
A is the final pupolation
P is the initial population 
e is the Euler's constant
r is the decay rate 
t is the time in years

A. We know for our problem that the initial population is 1,250, so P=1250; we also know that after a year the population is 1000, so A=1000 and t=1. Lets replace those values in our formula to find r:
A=Pe^{rt}
1000=1250e^{r}
e^{r}= \frac{1000}{1250}
e^{r}= \frac{4}{5}
ln(e^{r})=ln( \frac{4}{5} )
r=ln( \frac{4}{5} )
r=-02231

Now that we have r, we can write a function to model this scenario:
A(t)=1250e^{-0.2231t}.

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C. 
- The function is decreasing
- The function doe snot have a x-intercept 
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0: 
A(0)=1250e^{(-0.2231)(0)
A_{0}=1250e^{0}
A_{0}=1250
- Over the interval [0,10], the function will have a minimum at t=10:
A(10)=1250e^{(-0.2231)(10)
A_{10}=134.28

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: m= \frac{A(0)-A(10)}{10-0}
where 
m is the rate of change 
A(10) is the function evaluated at 10
A(0) is the function evaluated at 0
We know from previous calculations that A(10)=134.28 and A(0)=1250, so lets replace those values in our formula to find m:
m= \frac{134.28-1250}{10-0}
m= \frac{-1115.72}{10}
m=-111.572
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

7 0
3 years ago
Can someone help meeee pleaseeee!!!
inna [77]
I took the test it’s 29
4 0
3 years ago
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