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ELEN [110]
3 years ago
14

Would this be 18?? Please help

Mathematics
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

yes it would

Step-by-step explanation:

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What is the solution set of the following equation? -5x + x + 9 = -4x + 12
podryga [215]

A linear equation with one variable, x, is an equation that can be written in the standard form a x+b=0 where a and b are real numbers and a≠0. For example. 3x−12=0. A solution to a linear equation is ...Answer:

Step-by-step explanation:A linear equation with one variable, x, is an equation that can be written in the standard form a x+b=0 where a and b are real numbers and a≠0. For example. 3x−12=0. A solution to a linear equation is ...

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1) Find the mean for the given sample data. Unless indicated otherwise, round your answer to one more decimal place than is pres
Semmy [17]
Alright, so #1 is C, 4.88.
4 0
4 years ago
Read 2 more answers
What is the answer?
Dmitrij [34]

Answer:

read explanation

Step-by-step explanation:

rearrange equations to form y=mx+b

-2x+4=2y

-x+2=y

-y=-2x-5

y=2x+5

use these equations to find slope and y intercepts, and match those with the right graph

slope of first: -1

slope of second: 2

y-int of first: 2

y-int of second:5

3 0
3 years ago
At the beginning of a business day a banks vault held 575,900. by the end of the day 3.5 x 10 to the 3rd power was added. how mu
mr_godi [17]

Answer:

3.5*10^3=3,500

3,500+ 575,900=579,400

Step-by-step explanation:

7 0
3 years ago
shakira invests $500000 at 5% compound intwerest per annum . calculate how many years it takes for the value to double in value?
Kay [80]

Answer:

t=14.2\ years    

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=?\ years\\ P=\$500,000\\ r=0.05\\n=1\\ A=\$1,000,000  

substitute in the formula above  and solve for t

1,000,000=500,000(1+\frac{0.05}{1})^{(1)t}

2=(1.05})^{t}  

Apply log both sides

log(2)=log[(1.05})^{t}]  

log(2)=(t)log(1.05)  

t=log(2)/log(1.05)  

t=14.2\ years          

8 0
3 years ago
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