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3 years ago

(1+1/3)^2−2/9 I am about to have a mental breakdown, please somebody answer

Mathematics

2 answers:

Archy [21]3 years ago

8 0

Answer:

$\frac{14}{9} = 1\frac{5}{9} \\\\1.5$

Step-by-step explanation:

$(1+1/3)^2-2/9\\\\= \left(1+\frac{1}{3}\right)^2-\frac{2}{9}\\\\\left(1+\frac{1}{3}\right)^2=\frac{4^2}{3^2}\\\\=\frac{4^2}{3^2}-\frac{2}{9}\\\\\frac{4^2}{3^2}=\frac{16}{9}\\\\=\frac{16}{9}-\frac{2}{9}\\\\\mathrm{Apply\:rule}\:\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\\\=\frac{16-2}{9}\\\\\mathrm{Subtract\:the\:numbers:}\:16-2=14\\=\frac{14}{9}$

Andrej [43]3 years ago

3 0

Answer:

14/9

Step-by-step explanation:

take a look at my picture

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Can somebody help me please.

Fiesta28 [93]

I’m pretty sure it is 57?

7 0

2 years ago

5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

5 0

3 years ago

A prime number is a number that is evenly divisible only by 1 and itself. The prime numbers less than 100 are listed below.2 3 5

Usimov [2.4K]

Answer:

a. 0.96 = 96% probability that the number is odd.

b. 0.48 = 48% probability that the sum of the digits is odd.

c. 0.24 = 24% probability that the number is greater than 70.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

We are given the prime numbers less than 100, they are:

2,3,5,7,11,13,17,19,23,29,31,37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

There are 25 prime numbers less than 100.

a. The number is odd

Of the 25, 24 are odd, only the number 2 is even. So

24/25 = 0.96 = 96%

0.96 = 96% probability that the number is odd.

b. The sum of the digits is odd

The sum of the digits is odd for:

3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89. So for 12 of them

12/25 = 0.48 = 48%

0.48 = 48% probability that the sum of the digits is odd.

c. The number is greater than 70

71, 73, 79, 83, 89, 97. 6 of them

6/25 = 0.24 = 24%

0.24 = 24% probability that the number is greater than 70.

5 0

3 years ago

HELP PLEASE!!! The expected value of a random variable X is 35. The variable is transformed

Sunny_sXe [5.5K]

Answer:

Expected value x= 35

linear transformation is defined as a + bx

here, b=4, a=1

The transformation is $z=1+4x$

now, expected value, $l_z=l_z(a+bx)$

$=l(a)+l(bx)$

$=a+b\:l\:x$

substitute the value of a=1, b=4 and l=35

$l_z=1+4\times35$

$=1+140$

$=141$

So, the expected value of the transformed variable is <u>141</u>.

<u>OAmalOHopeO</u>

7 0

2 years ago

Read 2 more answers

Using distributive property what are two ways to find the product of 127 and 32?

timofeeve [1]

127 = 100 +27
32 = 30 +2
so 127 x 32 = (100 + 27) x (30+2) = (100x30) + (100x2) + (27x30) +(27x2)
= 3000+200+810+54 = 4064

second way:

127 = 100 + 27
127 x 32 = (100+27) x 32 = (100x 32) + ( 27x32) = 3200 + 864 = 4064

6 0

3 years ago