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vodka [1.7K]
3 years ago
12

Which statements about the trend line in item 3 are true? select all that apply.​

Mathematics
1 answer:
natita [175]3 years ago
5 0

Answer:

a and c

Step-by-step explanation:

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Anna wants to buy a car in two years and wants to have $10,000 for the purchase. How much will Anna need to deposit today in an
Brut [27]

Answer:

Anna will need to deposit \$9,050.25

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

A=P(1+\frac{r}{n})^{nt}  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

t=2\ years\\A=\$10,000\\ r=0.05\\n=12  

substitute in the formula above  and solve for P

10,000=P(1+\frac{0.05}{12})^{12*2}  

10,000=P(1.0042)^{24}  

P=10,000/(1.0042)^{24}    

P=\$9,050.25    

8 0
3 years ago
HELP will give branliest!
amm1812

Answer:

perimeter 140

Step-by-step explanation:

60L x 10W = 600

60(2) + 10(2) = 140

:D

8 0
3 years ago
What is the correct answer?
Tasya [4]

Given:

The function is

f(x)=(x+3)^2(x-5)^6

To find:

The zeros of the given function.

Solution:

The general form of polynomial is

P(x)=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}       ...(i)

where, a is a constant, c_1,c_2,...,c_n are zeros of respective multiplicities m_1,m_2,...,m_n.

We have,

f(x)=(x+3)^2(x-5)^6

On comparing this with (i), we get

c_1=-3,m_1=2

c_2=5,m_2=6

It means, -3 is a zero with multiplicity 2 and 5 is a zero with multiplicity 6.

Therefore, the correct option is B.

5 0
3 years ago
10 of the x power= 100000000​
zheka24 [161]

Answer:

x=8

Step-by-step explanation:

5 0
3 years ago
Help me on this to I'm begging u I'm so sorry I'm up so late but the school I'm in is wack ​
lianna [129]

Answer:

Answer below

Step-by-step explanation:

1. C 2

2. The temperature from 15°F to -6°F has changed by 21°F

5. C $37

I hope this helps

6 0
3 years ago
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