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Agata [3.3K]
2 years ago
7

I can't write the problem but I’ll insert in the picture

Mathematics
1 answer:
Hoochie [10]2 years ago
8 0
It’s 60 degrees
Ex: the total number when the degrees is added together should be 180
You might be interested in
HELP
Ket [755]

Given the radius of 50 miles and the line joining the cities at (0, 56) and (58, 0), the transmitter signal can be picked during 59.24 miles of the drive.

<h3>How can the duration of signal reception be found?</h3>

Radius of broadcast of the transmitter = 50 miles

Location of starting point = 56 miles north of the transmitter

Location of destination city = 58 miles east of the transmitter

Therefore we have;

Slope of the line joining the two cities

= 56 ÷ (-58) = -0.966

Which gives the equation of the line as follows;

y = -0.966•x + 56

The equation of the circle is;

{x}^{2}  +  {y}^{2}  =  {50}^{2}

{x}^{2}  +  { ( - 0.966x + 56)}^{2}  =  {50}^{2}

1.933156•x^2 - 108.192•x + 636 = 0

Which gives;

  • x = 6.67 or x = 49.29

Therefore;

When x = 6.67, we have;

  • y = -0.966 × 6.67 + 56 = 49.56

When x = 49.29, we have;

  • y = -0.966 × 49.29 + 56 = 8.4

The length of the drive, during which the driver can pick the signal, <em>l</em>, is therefore;

l = √((49.56-8.4)^2 + (49.29-6.67)^2) = <u>59.24 miles</u>

  • The length of the drive during which the signal is received is 59.24 miles

Learn more about the equation of a circle here:

brainly.com/question/502872

#SPJ1

5 0
2 years ago
HELPPP ITS QUESTION 17 BTW
Veronika [31]

Answer:

c

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
WEIGHT. An actor gains 20 pounds for a part and then loses 15 pounds during the filming of the movie to go along with the story.
MatroZZZ [7]
Okay, what’s the rest of the question?
3 0
3 years ago
Let z=3+i, <br>then find<br> a. Z²<br>b. |Z| <br>c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq
zysi [14]

Given <em>z</em> = 3 + <em>i</em>, right away we can find

(a) square

<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>

(b) modulus

|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(<em>z</em>) = arctan(1/3)

Then

<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))

<em>z</em> = √10 exp(<em>i</em> arctan(1/3))

or

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

(c) square root

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

√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)

and

√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 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 <em>z</em> lies in the first quadrant, so

0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2

which means

0 < 1/2 arctan(1/3) < <em>π</em>/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(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),

\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 <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 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 <em>z</em> 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
Renee is sewing a quilt whose pattern contains right triangles.
Mashutka [201]

Answer:

  8 inches

Step-by-step explanation:

The area of a triangle is ...

  A = 1/2bh

Filling in the given values, we have ...

  24 = 1/2b(6)

  24/3 = b = 8 . . . . . divide by the coefficient of b

The base of each triangular quilt piece is 8 inches long.

7 0
3 years ago
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