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3 years ago

What is required to name a plane

1 answer:

3 0

You need three non-collinear points to name a plane.

Non-collinear means "not on the same straight line". So these three points can't all lie on the same straight line. If three points do lie on the same straight line, then it is impossible to generate a single unique plane.

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What is the simpest form of 69 over 30?

avanturin [10]

2.3 in decimal form

4 0

3 years ago

In what time the compound amount of Rs 60000 at 10% rate is rs 79860?

Ksenya-84 [330]

Answer:

P= Rs 60000

A= Rs 79860

T=1 & 1/2 year = 3/2 years

= 3/2 x 2 = 3 half years

R= ?

Applying the formula A= P (1+r/100)^T

79860 = 60000 (1+ r/100)^3

79860/60000 = (1+r/100)^3

1331/1000 = (1+r/100)^3

root(3)(1331/1000) = (1+r/100)

11/10 = 1+r/100

11/10 -1 = r/100

1/10 = r/100

r= 10 %

Step-by-step explanation:

8 0

3 years ago

NO SCAM. PLEASE SHOW WORK. WILL GIVE BRAINLY TO CORRECT ANSWER.

sergeinik [125]

Answer:

$140$

Step-by-step explanation:

A regular polygon is a shape with equal sides and angles. Because of this, we can write the following equation to set two sides equal to each other:

$3a+b+3=2a+2b+6$

Solving, we will get a in terms of b:

$3a=2a+b+3,\\a=b+3$

Now we can substitute a for (b+3) into our equation:

$5a-2b+1=2a+2b+6,\\5(b+3)-2b+1=2(b+3)+2b+6,\\3b+16=4b+12,\\-b=-4,\\b=\boxed{4}$

Therefore, the length of each side of this polygon is $5(4+3)-2(4)+1=28$ .

Since the perimeter of the polygon consists of all five sides, the perimeter is:

$p=28\cdot 5 =\boxed{140}$

6 0

3 years ago

Read 2 more answers

15 POINTS!<br> What is the correct answer for Dean's problem? Explain where he went wrong.

prohojiy [21]

Answer:

The slope is 5/-2.

Step-by-step explanation:

Slope is y1-y2 over m1-m2 (rise over run). The first ordered pair is -2 (m1) and 11 (y1). We then subtract the second ordered pair (4 (m2) and -4 (y2)) from the first.

11 - (-4) = 11 + 4 = 15

-2 - 4 = -6

Remember, slope is rise over run (y over x), so the slope is 15/-6. Now, we must simplify. 15/-6 = 5/-2

Dean went wrong because he thought that slope was run over rise (x over y). If he had switched the two numbers, his answer would have been correct.

8 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