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3 years ago

Which is the most accurate measurement of 1 mile? (Remember, there are

Mathematics

1 answer:

olasank [31]3 years ago

4 0

Bro bruhhhh bruh bruh ahahahhaah 2092&29 answer is D

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Write each complex number in rectangular form. a. 2(cos(135)+i sin(135)) b. 3(cos(120))+i sin(120)) c. 5(cos(5pi/4)+i sin(5pi/4)

bagirrra123 [75]

$\bf r\left[ cos\left( \theta \right)+i\ sin\left( \theta \right) \right]\quad 
\begin{cases}
x=rcos(\theta )\\
y=rsin(\theta )
\end{cases}\implies 
\begin{array}{llll}
x&,&y\\
a&&b
\end{array}\implies a+bi\\\\
-------------------------------\\\\
2\left[ cos\left( 135^o\right)+i\ sin\left( 135^o\right) \right]\impliedby r=2\qquad \theta =135^o
\\\\\\
2\left( -\frac{\sqrt{2}}{2} \right)+i\ 2\left( \frac{\sqrt{2}}{2}\right)\implies -\sqrt{2}+\sqrt{2}\ i$

$\bf -------------------------------\\\\
3\left[ cos\left( 120^o\right)+i\ sin\left( 120^o\right) \right]\impliedby r=3\qquad \theta =120^o
\\\\\\
3\left( -\frac{1}{2} \right)+i\ 3\left( \frac{\sqrt{3}}{2}\right)\implies -\frac{3}{2}+\frac{3\sqrt{3}}{2}\ i$

$\bf \\\\
-------------------------------\\\\
5\left[ cos\left( \frac{5\pi }{4}\right)+i\ sin\left( \frac{5\pi }{4}\right) \right]\impliedby r=5\qquad \theta =\frac{5\pi }{4}
\\\\\\
5\left( -\frac{\sqrt{2}}{2} \right)+i\ 5\left( -\frac{\sqrt{2}}{2}\right)\implies -\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\ i$

$\bf -------------------------------\\\\
4\left[ cos\left( \frac{5\pi }{3}\right)+i\ sin\left( \frac{5\pi }{3}\right) \right]\impliedby r=4\qquad \theta =\frac{5\pi }{3}
\\\\\\
4\left( \frac{1}{2} \right)+i\ 4\left( -\frac{\sqrt{3}}{2}\right)\implies \frac{1}{2}-\frac{\sqrt{3}}{2}\ i$

3 0

3 years ago

Witch of the following are the factors of 4b^2 - 16?

luda_lava [24]

4b^2 - 16 = (2b + 4)(2b - 4)

difference of squares
a^2 – b^2 = (a + b)(a – b)

4 0

3 years ago

Find the area of the shaded region. Round the the nearest tenth.

Brums [2.3K]

Answer:

226.084 sq.m.

Step-by-step explanation:

Radius of circle = 18.6 cm

Area of sector = $\frac{\theta}{360^{\circ} }\pi r^2$

Area of sector = $\frac{123}{360} \times 3.14 \times (18.6)^2$

Area of sector = $371.157 m^2$

Area of triangle = $\frac{1}{2} \times a \times b \times Sin x$

a = OA = radius = 18.6 cm

b = OB = radius = 18.6 cm

x =123 °

Substitute the values in the formula

Area of triangle = $\frac{1}{2} \times 18.6 \times 18.6 \times Sin 123$

Area of triangle = $145.073 m^2$

So, Area of shaded region = Area of sector - Area of triangle

= $371.157 m^2-145.073 m^2$

= $226.084 m^2$

Hence the area of the shaded region is 226.084 sq.m.

5 0

3 years ago

A telephone pole breaks and falls as shown. To the nearest foot, what was the original height of the pole? Please i need this as

8090 [49]

I don’t know how old this is but hopefully I’m not too late it’s 17 7+10=17 it is NOT 70 ‍♀️

7 0

3 years ago

1) Solve the following system of equations algebraically. Give your solution as two

Bad White [126]

For this case we have the following system of equations:

$y = x ^ 2 + 7x + 12\\y = 2x + 8$

Equating both equations we have:

$x ^ 2 + 7x + 12 = 2x + 8\\x ^ 2 + 7x-2x + 12-8 = 0\\x ^ 2 + 5x + 4 = 0$

We must find the solutions, for this we factor. We look for two numbers that, when multiplied, result in 4 and when added, result in 5. These numbers are 4 and 1:

$4 + 1 = 5\\4 * 1 = 4$