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maks197457 [2]
3 years ago
11

I need help which triangle theorem do I use for these

Mathematics
1 answer:
irina [24]3 years ago
4 0

Answer:

Step-by-step explanation: B. SSS because it shows us three sides ,C.RHS because it has a right angle, d. AAS because it shows us two sides and a non included side,  e. RHS because it has a right angle,  f. AAS because it has two angles and non included side.

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What is the coefficient of the term x5y7 in the expansion of (x - y)12 ?
lozanna [386]

Answer:

12 and -12

Step-by-step explanation:

12x-12y is the expansion using the distributive property, so the coefficient of x is 12 and the coefficient of y is -12

6 0
3 years ago
Find the fourth roots of 16(cos 200° + i sin 200°).
NeTakaya

Answer:

<em>See below.</em>

Step-by-step explanation:

To find roots of an equation, we use this formula:

z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos(\frac{\theta}{n}+\frac{2k\pi}{n} )+\mathfrak{i}(sin(\frac{\theta}{n}+\frac{2k\pi}{n})), where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the amount of roots needed - 0 is included).

In this case, n = 4.

Therefore, we adjust the polar equation we are given and modify it to be solved for the roots.

Part 2: Solving for root #1

To solve for root #1, make k = 0 and substitute all values into the equation. On the second step, convert the measure in degrees to the measure in radians by multiplying the degrees measurement by \frac{\pi}{180} and simplify.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(0)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(0)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))

z^{\frac{1}{4}} = 2(sin(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))

<u>Root #1:</u>

\large\boxed{z^\frac{1}{4}=2(cos(\frac{19\pi}{36}))+\mathfrack{i}(sin(\frac{19\pi}{38}))}

Part 3: Solving for root #2

To solve for root #2, follow the same simplifying steps above but change <em>k</em>  to k = 1.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(1)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(1)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{2\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{2\pi}{4}))\\

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{2}))\\

<u>Root #2:</u>

\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{7\pi}{9}))+\mathfrak{i}(sin(\frac{7\pi}{9}))}

Part 4: Solving for root #3

To solve for root #3, follow the same simplifying steps above but change <em>k</em> to k = 2.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(2)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(2)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{4\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{4\pi}{4}))\\

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\pi))+\mathfrak{i}(sin(\frac{5\pi}{18}+\pi))\\

<u>Root #3</u>:

\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{23\pi}{18}))+\mathfrak{i}(sin(\frac{23\pi}{18}))}

Part 4: Solving for root #4

To solve for root #4, follow the same simplifying steps above but change <em>k</em> to k = 3.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(3)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(3)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{6\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{6\pi}{4}))\\

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{3\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{3\pi}{2}))\\

<u>Root #4</u>:

\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{16\pi}{9}))+\mathfrak{i}(sin(\frac{16\pi}{19}))}

The fourth roots of <em>16(cos 200° + i(sin 200°) </em>are listed above.

3 0
3 years ago
Solve using elimination method
Anon25 [30]
X+y=8
2x^2-y=-5
elimination method means you link the 2 equations together to make one
these two already have y in the perfect form where you can cancel it out +y and -y so no need to perform any other action to a certain equation.. all you need to do is add them now
2x^2+x=3
7 0
3 years ago
Help please?
Nikolay [14]

Answer:

An angle bisector.

Step-by-step explanation:

This is a ray (Line that extends only one infinitely while the other way has an eventual and perceivable end) that divides an angle (in this case, the angle of a triangle) into two equal parts.

7 0
3 years ago
Hat is the equation of the quadratic function with a vertex at (2,–25) and an x-intercept at (7,0)?
Novosadov [1.4K]

Answer:

y = x² - 4x - 21

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Here (h, k) = (2, - 25), thus

y = a(x - 2)² - 25

To find a substitute (7, 0) into the equation

0 = a(7 - 2)² - 25 = a(5)² - 25 = 25a - 25 ( add 25 from both sides )

25a = 25 ( divide both sides by 25

a = 1, thus

y = (x - 2)² - 25 ← in vertex form

Expand and simplify

y = x² - 4x + 4 - 25

y = x² - 4x - 21 ← in standard form

8 0
3 years ago
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