How many solutions does the system of equations have y=5/2x+2 one zero two

true or false? The sum of the difference is not always zero for any distribution consisting of n observations

mezya [45]

Answer:

i believe the answer is false !

Step-by-step explanation:

3 0

2 years ago

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.

7 0

4 years ago

Factor completely 56a2b3 − 35ab.

VMariaS [17]

I would start off by taking away 1a. That would make the problem be 56ab3-35b.I only took away 1 because each have at least 1a and is okay to do.

Next I would deal with the variable b. I would cross of 1 b. That's because both sides have at least 1b. Now, it's shortened to be 56ab2-35.

Since you cannot take away anymore variables, you have to deal with 56 and 35. I start small with dividing each by 2. I am trying to see what the greatest number could be while making the numbers still be whole. That turns 56 into 28 when it's cut in half. The 35 now turns into 17.5.

I would assume your teacher would want the numbers to be whole. seeing as though when 35 is cut in half and makes a decimal number, I would leave them. What I mean by that is to leave the numbers as 56 and 35.

So, that means the answer is 56ab2-35.

I hope this helps!! (And makes sense)

5 0

4 years ago

(!!!!PLEASE HELP ASAP!!!!!) Graph the first room using the coordinates: (1, 1), (1, 7), (3, 1) and (3, 7).

Dima020 [189]

Answer:

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-negative-number-topic/cc-6th-coordinate-plane/a/graphing-coordinates-review

Graphing coordinates review (article) - Khan Academy

4 0

2 years ago

4x-5y+2x+4+6+9y+7x+2