All categories

3 years ago

Which number, when placed in the box, will make the equation true?

1 answer:

8 0

$\mathsf{\heartsuit\;\; (\sqrt{8})^2 = \sqrt{64}}\\\\\mathsf{\heartsuit\;\; [(8)^\frac{1}{2}]^2 = \sqrt{(8)^2}}\\\\\mathsf{\heartsuit\;\; (8)^\frac{2}{2} = [(8)^2]^\frac{1}{2}}\\\\\mathsf{\heartsuit\;\; 8 = [8]^\frac{2}{2}}\\\\\mathsf{\heartsuit\;\; 8 = 8}\\\\\mathsf{\heartsuit\;\; Option\;(A)\;is\;the\;Answer}$

You might be interested in

Varies inversely as If x = 6 then y = 8. Find y when x = 4 .<br> Y=

icang [17]

Answer:

Explanation:

The statement is expressed as

∝

inverse means

variable

To convert to an equation introduce k, the constant of variation.

⇒

To find k use the given condition that y = 8 when x = 4

⇒

equation is

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

∣

−−−−−−−−−−−−

→

⇒

when

8 0

3 years ago

Please help me this is for a grade

IgorC [24]

Answer:

Step-by-step explanation:

4 0

3 years ago

Which of the following is not one of the 8th roots of unity?

Anika [276]

Answer:

1+i

Step-by-step explanation:

To find the 8th roots of unity, you have to find the trigonometric form of unity.

1. Since $z=1=1+0\cdot i,$ then

$Rez=1,\\ \\Im z=0$

and

$|z|=\sqrt{1^2+0^2}=1,\\ \\\\\cos\varphi =\dfrac{Rez}{|z|}=\dfrac{1}{1}=1,\\ \\\sin\varphi =\dfrac{Imz}{|z|}=\dfrac{0}{1}=0.$

This gives you $\varphi=0.$

Thus,

$z=1\cdot(\cos 0+i\sin 0).$

2. The 8th roots can be calculated using following formula:

$\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.$

Now

at k=0, $z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;$

at k=1, $z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=2, $z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;$

at k=3, $z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};$

at k=4, $z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;$

at k=5, $z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

at k=6, $z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;$

at k=7, $z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};$

The 8th roots are

$\{1,\ \dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ i, -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2},\ -1, -\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2},\ -i,\ \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2}\}.$

Option C is icncorrect.

5 0

3 years ago

Find the area of the shape. Either enter an exact answer in terms of π or use 3.14 for π and enter your answer as a decimal. (AN

olga2289 [7]

3/4 * pi * 16 = 12pi

3 0

3 years ago

Okay, can someone private message me and help me with some math problems pleaseeeeeee????? Thank you!

Marizza181 [45]

I can help you, as long as it's something I've already learned so as long as it anything less than stats. I'll pm you.

7 0

3 years ago