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Serggg [28]
3 years ago
14

Choose the equation of the horizontal line that passes through the part point (1,-5)

Mathematics
2 answers:
vladimir2022 [97]3 years ago
4 0
We know is a horizontal line, so, if it passes through 1,-5, it also passes through "whatever", -5, like 20, -5 or 1000000, -5, or -100000000, -5 and so on.

so, let's pick another point say -7, -5, check the picture below, and let's check about the equation that runs through it,

\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ 1 &,& -5~) 
%  (c,d)
&&(~ -7 &,& -5~)
\end{array}
\\\\\\
% slope  = m
slope =  m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-5-(-5)}{-7-1}
\\\\\\
\cfrac{-5+5}{-7-1}\implies \cfrac{0}{-8}\implies 0
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-5)=0(x-1)
\\\\\\
y+5=0\implies y=-5

Natalija [7]3 years ago
3 0
The answer is A. Y=-5
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then,

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If 10000 is invested at an interest rate of 10 per year ,compound semiannually,find the value of the investment after the given
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Answer:

Part a) \$17,958.56  

Part b) \$32,251.00  

Part c) \$57,918.16

Step-by-step explanation:

<u><em>The complete question is</em></u>

If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, find the value of the investment after the given number of years. (Round your answers to the nearest cent.)

a)6 years

b)12 years

c)18 years

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

Part a) 6 years

we have  

t=6\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*6}  

A=10,000(1.05)^{12}  

A=\$17,958.56  

Part b) 12 years

we have  

t=12\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*12}  

A=10,000(1.05)^{24}  

A=\$32,251.00  

Part c) 18 years

we have  

t=18\ years\\ P=\$10,000\\ r=10\%=10/100=0.10\\n=2  

substitute in the formula above  

A=10,000(1+\frac{0.10}{2})^{2*18}  

A=10,000(1.05)^{36}  

A=\$57,918.16

7 0
4 years ago
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