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3 years ago

Evaluate. 418! 2!7! plzzzz helpppp

Mathematics

2 answers:

zavuch27 [327]3 years ago

7 0

Answer:

whatttttt6 whattttt whattt

ZanzabumX [31]3 years ago

7 0

59.7

Step-by-step explanation:

You might be interested in

Please answer both!

Naya [18.7K]

Answer:

36)12(37)0.67

Step-by-step explanation:

36)A=base×height

A=2×6

A=12

37)A=a+b/2h

A=2+6/12

A=8/12

A=0.67

8 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

Marco does 14 extra math problems each school night how many extra problems does he do each week. there are 5 school nights in a

Alexus [3.1K]

Marco does 70 extra math problems each week. 14(5)=70

4 0

3 years ago

Read 2 more answers

One of the giraffes, Bernard, is sick. Last week he weighed 2,490 pounds. This week he weighs 2,312 pounds. If w represents the

victus00 [196]

Answer:

$w+2,312=2,490$

$w=178\ pounds$

Step-by-step explanation:

Let

w ----> the amount of weight lost in pounds

we know that

The amount of weight Bernard lost plus the weight of this week must be equal to the weight of the last week

The linear equation that represent this situation is

$w+2,312=2,490$

solve for w

subtract 2,312 both sides

$w=2,490- 2,312=178\ pounds$

4 0

3 years ago

Read 2 more answers

The Polygons below are similar. Find the value of z. A. 4.5 B. 7.5 C. 12 D. 16

Misha Larkins [42]

The correct answer is B. 7.5
Similar polygons are polygons wherein its corresponding sides are proportional. All corresponding sides of the larger polygon are proportional to the corresponding sides to the smaller polygon, wherein segment HG is proportional to segment DC, and segment FG is proportional to segment BC. (See attachment for reference) A proportion is an equation that states two ratios are equivalent. By cross multiplying the two ratios, you get an equation for finding the value of z, which is 7.5

8 0

4 years ago