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2 years ago

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

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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, l, is therefore;

l = √((49.56-8.4)^2 + (49.29-6.67)^2) = 59.24 miles

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:

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, 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

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