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3 years ago

Is ( 0,0) a solution to this system? y<=x^2-4 y>2x-1?

Mathematics

2 answers:

pashok25 [27]3 years ago

7 0

No.

Your proposed solution does not satisfy the first inequatlity.
0 ≤ 0 - 4 . . . . . not true

Xelga [282]3 years ago

3 0

Answer:

The point $(0,0)$ is not a solution of the system of inequalities

Step-by-step explanation:

we have

$y\leq x^{2} -4$ -----> inequality A

$y> 2x-1$ -----> inequality B

we know that

If a ordered pair is a solution of the system of inequalities

then

the ordered pair must be satisfy the inequalities of the system

Verify

For $x=0, y=0$

substitute the value of x and the value of y in the inequalkity A and in the inequality B

Inequality A

$0\leq 0^{2} -4$

$0\leq -4$ -------> is not true

therefore

The point $(0,0)$ is not a solution of the system of inequalities

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If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600? If the co

Sphinxa [80]

Answer:

Using continuous interest 6.83 years before she has $1600.

Using continuous compounding, 6.71 years.

Step-by-step explanation:

Compound interest:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

Continuous compounding:

The amount of money earned after t years in continuous interest is given by:

$P(t) = P(0)e^{rt}$

In which P(0) is the initial investment and r is the interest rate, as a decimal.

If Tanisha has $1000 to invest at 7% per annum compounded semiannually, how long will it be before she has $1600?

We have to find t for which $A(t) = 1600$ when $P = 1000, r = 0.07, n = 2$

$A(t) = P(1 + \frac{r}{n})^{nt}$

$1600 = 1000(1 + \frac{0.07}{2})^{2t}$

$(1.035)^{2t} = \frac{1600}{1000}$

$(1.035)^{2t} = 1.6$

$\log{1.035)^{2t}} = \log{1.6}$

$2t\log{1.035} = \log{1.6}$

$t = \frac{\log{1.6}}{2\log{1.035}}$

$t = 6.83$

Using continuous interest 6.83 years before she has $1600

If the compounding is continuous, how long will it be?

We have that $P(0) = 1000, r = 0.07$

Then

$P(t) = P(0)e^{rt}$

$1600 = 1000e^{0.07t}$

$e^{0.07t} = 1.6$

$\ln{e^{0.07t}} = \ln{1.6}$

$0.07t = \ln{1.6}$

$t = \frac{\ln{1.6}}{0.07}$

$t = 6.71$

Using continuous compounding, 6.71 years.

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