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4 years ago

Help please.........

Mathematics

1 answer:

murzikaleks [220]4 years ago

5 0

The answer is x = 75 which is found through solving x+105 = 180 (subtract 105 from both sides)

Any time there's an inscribed quadrilateral like this, the opposite angles always add to 180 degrees (opposite angles are supplementary).

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A square has a perimeter of 25 inches. Find the length of a side.

Arada [10]

Answer: The area is a = 6.25

Step-by-step explanation: Using formula P = 4 a

Solving for a

a = $\frac{p}{4}$ = $\frac{25}{4}$ = 6.25 in

4 0

3 years ago

Find the cube roots of 27(cos 330° + i sin 330°)

Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

DeMoivre's Theorem

Let $z=r(cos\theta+isin\theta)$ be a complex number in polar form, where $n$ is an integer and $n\geq1$ . If $z^n=r^n(cos\theta+isin\theta)^n$ , then $z^n=r^n(cos(n\theta)+isin(n\theta))$ .

Nth Root of a Complex Number

If $n$ is any positive integer, the nth roots of $z=rcis\theta$ are given by $\sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}}$ where the nth roots are found with the formulas:

$\sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr]$ for degrees (the one applicable to this problem)
$\sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr]$ for radians

for $k=0,1,2,...\:,n-1$

Calculation

$z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)$

First cube root where k=2

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]$

Second cube root where k=1

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]$

Third cube root where k=0

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]$

4 0

2 years ago

Find the surface area. Show work.

kotegsom [21]

Step-by-step explanation:

first you'll have to cut the triangle in half to make the right angle. when u do the side is whatever the num is 132-??= thats the answer. and then add all of the area nums and then u get the num

7 0

3 years ago

Read 2 more answers

Factor the expression completely. which is a factor of the given expression?xy + y +3x + 3a) (y + 3)

aksik [14]

xy + y +3x + 3
y(x + 1) + 3(x + 1)
(y + 3)(x + 1)

Factors: (y+3) and (x+1)

7 0

4 years ago

Find radiuses of circles, if OO1=20cm and 3AK=2BK

Serggg [28]

Answer:

radius of smaller circle on the left = 8 centimeters
radius of larger circle on the right = 12 centimeters

=============================================================

Explanation:

The letter O looks very similar to the number zero, so I'm going to use other letters to avoid confusion. For the smaller circle on the left, it'll have center P. The circle on the right will have center Q. See the diagram below.

Based on that, we can say that PQ = 20 cm.

Let x = length of PK

PK + KQ = PQ

KQ = PQ - PK

KQ = 20 - x

We'll keep this in mind for later. I'll refer to this as "Fact 1" when it does come up later.

------------------------

We're given this equation

3AK=2BK

Let's get the numbers to one side and the letters to the other side like so:

3AK=2BK

3AK/BK = 2

AK/BK = 2/3

We'll use this later as well. I'll refer to this as "Fact 2" when it does come up later.

------------------------

Next, draw a segment from P to A to form triangle PAK (also shown in the diagram below). Do the same for points Q and B to form triangle QBK.

The segments PA and PK are radii of the smaller circle, so PA = PK which leads to triangle PAK being isosceles. Triangle QBK is also isosceles for similar reasoning.

The angles AKP and QKB are vertical angles, so they are congruent.

The facts mentioned so far boil down to the two triangles being similar triangles (use the angle angle method to prove).

------------------------

Now that we know the triangles are similar, we can use a proportion to solve for x.

PK/KQ = AK/KB

PK/KQ = 2/3 .... use fact 2

x/(20-x) = 2/3 .... use fact 1

3x = 2(20-x) .... cross multiply

3x = 40 - 2x

3x+2x = 40

5x = 40

x = 40/5

x = 8

So,

PK = x = 8
KQ = 20-x = 20-8 = 12

The radius of the smaller circle on the left is 8 cm.

The radius of the larger circle on the right is 12 cm.

4 0

2 years ago