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Citrus2011 [14]
3 years ago
7

Rebecca draws two triangles on this coordinate plane.

Mathematics
1 answer:
anzhelika [568]3 years ago
8 0

Answer:

The answer is below

Step-by-step explanation:

From the image attached, the coordinates of triangle ABC are at A(-5, 4), B(-5,7) and C(-2, 7)

a) If a point O(x,y) is rotated 90° clockwise, the new coordinate is O'(y, -x)

If triangle ABC is rotated 90° clockwise, its new coordinate is at:

A'(4, 5), B'(7, 5) and C'(7, 2)

The coordinate of A"B"C" is A"(2,5), B"(5,5), C(5,2)

Hence the transformation used to map A'(4, 5), B'(7, 5) and C'(7, 2) to A"(2,5), B"(5,5), C(5,2) is (x - 2, y + 0). Hence comparing with (x + h, y + k) gives:

h = -2, k = 0

b) If a point O(x,y) is reflected across the y = -x line, the new coordinate is O'(-y, -x)

If triangle ABC is reflected across y = -x, its new coordinate is at:

A'(-4, 5), B'(-7, 5) and C'(-7, 2)

The coordinate of A"B"C" is A"(2,5), B"(5,5), C(5,2)

You might be interested in
What is the solution set to the equation 4x2 – 16 = 0?
Luda [366]

Answer:

2

Step-by-step explanation:

add 16 to zero

4x2=16

Divide 4x by 2 and 16 by two

4x=8

divide 8 by 4

x=2

4 0
3 years ago
Read 2 more answers
What is the domain of the function<br> Y=2/x-5
aksik [14]

Answer:

x ≥ 5

Step-by-step explanation:

Find the domain by finding where the equation is defined.

Interval Notation:

(5, ∞ )

Set-Builder Notation:

{x | x ≥ 5}

So, the answer is : x ≥ 5

(PLEASE MARK ME AS BRAINLIEST!!!)

3 0
2 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
during a 3 day event a total of 7,458 people attended. if the same people of people attended each day, how many people attended
s344n2d4d5 [400]
Answer... I don't know if this is correct but here it is.

1864.5

Hope this helps! Brainliest would be appreciated!
4 0
3 years ago
Read 2 more answers
The game of American roulette involves spinning a wheel with 38 slots: 18 red, 18 black, and 2 green. A ball is spun onto the wh
solong [7]

Answer:

a)- \frac{1}{19}

b) SD(x) =  0.9986

Step-by-step explanation:

Given data:

Number of slots 38

slots for red ball 18

slots for black = 18

slots for green 2

outcomes       Red      Black or Green

Profit                  1                 -1

P(X)                   18/38       20/38

Expected value of wining,  

E(x) = 1. \frac{18}{38} + (-1) \frac{20}{38} = - \frac{2}{38} = - \frac{1}{19}

B) Standard deviation of winning SD(x)

SD(x) = \sqrt{(1- (-\frac{1}{19})^2 .\frac{18}{38} + (1- (-\frac{1}{19})^2 .\frac{20}{38}}

SD(x) = \sqrt{\frac{360}{361}

SD(x) =  0.9986

3 0
3 years ago
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