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3 years ago

PLZZZ HELP IM IN A RUSH

Mathematics

1 answer:

cupoosta [38]3 years ago

4 0

1/3 - 2/9 - 1/3

= -2/9
(Decimal: -0.222222)

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Kerion had a beaded business.she can make 12 necklaces in 2 hours.how long will it take her to make 9 necklaces

sammy [17]

It is 1 hour and 30 minutes

5 0

3 years ago

Express the fifth roots of unity in standard form a + bi. with 1 + 0i

marishachu [46]

Answer:

For K=0

$cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0$

For K=1:

$cos(\frac{0+2\pi 1}{5})+isin(\frac{0+2\pi 1}{5})\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510$

For K=2:

$cos(\frac{0+2\pi 2}{5} )+isin(\frac{0+2\pi 2}{5} )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587$

For K=3:

$cos(\frac{0+2\pi 3}{5} )+isin(\frac{0+2\pi 3}{5} )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587$

For K=4:

$cos(\frac{0+2\pi 4}{5} )+isin(\frac{0+2\pi 4}{5} )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510$

Step-by-step explanation:

Fifth Root is given by:

$\sqrt[5]{z}=1+0i$

The above equation will become:

$z=(1+0i)^5$

It can be written as:

$z=[cos(0)+isin(0)]^5$

|z|=1,

According to De-moivre's Theorem:

$z=cos(\frac{0}{5})+isin(\frac{0}{5})\\ z=cos(0)+isin(0)$

Now, Fifth Roots of unity in standard form a + bi :

$\sqrt[5]{z}=[{cos(0+2\pi k)+isin(0+2\pi k)}]^{1/5}$

k=0,1,2,3,4

For K=0

$cos(\frac{0+2\pi 0}{5})+isin(\frac{0+2\pi 0}{5})\\cos(0)+isin(0)=1+i0$

For K=1:

$cos(\frac{0+2\pi 1}{5})+isin(\frac{0+2\pi 1}{5})\\cos(2\pi/5)+isin(2\pi/5)=0.3090+i0.9510$

For K=2:

$cos(\frac{0+2\pi 2}{5} )+isin(\frac{0+2\pi 2}{5} )\\cos(4\pi/5)+isin(4\pi/5)=-0.809+i0.587$

For K=3:

$cos(\frac{0+2\pi 3}{5} )+isin(\frac{0+2\pi 3}{5} )\\cos(6\pi/5)+isin(6\pi/5)=-0.809-i0.587$

For K=4:

$cos(\frac{0+2\pi 4}{5} )+isin(\frac{0+2\pi 4}{5} )\\cos(8\pi/5)+isin(8\pi/5)=0.3091-i0.9510$

6 0

3 years ago

A giant red oak's diameter was 248 inches in 1965. The tree's diameter had grown to 251 inches in 2005. Find the average rate of

Anna71 [15]

Sorry for the late response

(bigger number - smaller number) divided by original number

(251 - 248) / 248
3 / 248 Divide to make this a decimal
0.0120967741... round it
0.012 change into percent
The rate of change is 1.2%

hope this helps!

4 0

4 years ago

Select each statement that is true for all polynomials

Sergio [31]

Answer: When two polynomials are multiplied, the product is always a

polynomial.

Step-by-step explanation:

The only true property for polynomials is the third options because when you do multiply polynomials, you can only get a polynomial.

For instance;

Multiply (5x – 6y) by (7x – 4y)

= (5x – 6y)(7x – 4y)

= 35x² - 20xy - 42xy + 24y²

= 35x² - 62xy + 24y²

The result is a polynomial.

The rest of the options will not always yield polynomials.

3 0

4 years ago

A cake is removed from a 310°F oven and placed on a cooling rack in a 72°F room. After 30 minutes the cake's temperature is 220°

Fynjy0 [20]

Answer:

The time is 135 min.

Step-by-step explanation:

For this situation we are going to use Newton's Law of Cooling.

Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium and is given by

$T(t)=C+(T_0-C)e^{kt}$

where,

C = surrounding temp

$T(t)$ = temp at any given time

t = time

$T_0$ = initial temp of the heated object

k = constant

From the information given we know that:

Initial temp of the cake is 310 °F.
The surrounding temp is 72 °F.
After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

$220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}$

$T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}$

Next, we find the time at which the cake's temperature will be 100°F.

$100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1$

4 0

3 years ago