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Shtirlitz [24]
3 years ago
14

The vertices of a triangle are P(−3, −4), Q(3, 4), and R(−6, −3). Name the vertices of Rx = 0 (PQR).

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
6 0
The correct answer is <span>A) P'(3, −4), Q'(−3, 4), R'(6, −3)</span>

Rx = 0 indicates a reflection over the y-axis. 

The rule for such a transformation is:
(x, y) --> (-x, y)
which means that the x-coordinate changes sign and the y-coordinate stays the same.

Therefore:
P<span>(-3, -4) --> P'(3, -4)
Q(3, 4) --> Q'(-3, 4)
R(-6, -3)</span> --> R'(6, -3)

These points are those in option A).
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Delicious77 [7]
5*7*3
35*3= 105
105 different ways
8 0
3 years ago
HELP ASAP WILL GIVE BRAINLY
Mila [183]

Answer:

1 and 2 are not polyhedrons, as a circle has infinite sides

1. cone, 1 vertex

2. sphere, nothing

3. pentagonal prism?

wait but can't every face be a base???

anyway 7 faces, 10 vertices

4. triangular prism

5 faces, 6 vertices

Step-by-step explanation:

8 0
3 years ago
In the accompanying diagram, line n is parallel to
mote1985 [20]
Here too same procedure
4) 156 is the answer
4 0
3 years ago
Express 80 inches in standard notation using feet and inches.
Vladimir79 [104]

80 inches in standard notation using feet and inches would be expressed as 6 ft 8 inches by converting inches into feet and inches.

The solution to the given problem is to use some standard conversion units that are:

  • 1 foot = 12 inches
  • 1 inch = 0.8333 feet

Solution:

As mentioned above that one inch is equal to 0.8333 foot therefore

1 foot = 12 inches

then,

80 inches would be equal to

= \frac{80}{12} ft

= \frac{20}{3} ft

= 6ft 8 inches

= 6' 8"

Thus, 80 inches in standard notation using feet and inches would be expressed as 6 ft 8 inches by converting inches into feet and inches.

Learn more:

brainly.com/question/884268

3 0
3 years ago
2. Find the general relation of the equation cos3A+cos5A=0
mars1129 [50]
<h2>Answer:</h2>

A=\frac{\pi}{8}+\frac{n\pi}{4}or\ A=\frac{\pi}{2}+n\pi

<h2>Step-by-step explanation:</h2>

<h3>Find angles</h3>

cos3A+cos5A=0

________________________________________________________

<h3>Transform the expression using the sum-to-product formula</h3>

2cos(\frac{3A+5A}{2})cos(\frac{3A-5A}{2})=0

________________________________________________________

<h3>Combine like terms</h3>

2cos(\frac{8A}{2})cos(\frac{3A-5A}{2})=0\\\\  2cos(\frac{8A}{2})cos(\frac{-2A}{2})=0

________________________________________________________

<h3>Divide both sides of the equation by the coefficient of variable</h3>

cos(\frac{8A}{2})cos(\frac{-2A}{2})=0

________________________________________________________

<h3>Apply zero product property that at least one factor is zero</h3>

cos(\frac{8A}{2})=0\ or\ cos(\frac{-2A}{2})=0

________________________________________________________

<h2>Cos (8A/2) = 0:</h2>

<h3>Cross out the common factor</h3>

cos\ 4A=0

________________________________________________________

<h3>Solve the trigonometric equation to find a particular solution</h3>

4A=\frac{\pi}{2}or\ 4A=\frac{3\pi}{2}

________________________________________________________

<h3>Solve the trigonometric equation to find a general solution</h3>

4A=\frac{\pi}{2}+2n\pi \ or\\ \\ 4A=\frac{3 \pi}{2}+2n \pi\\ \\A=\frac{\pi}{8}+\frac{n \pi}{4\\}

________________________________________________________

<h2>cos(-2A/2) = 0</h2>

<h3>Reduce the fraction</h3>

cos(-A)=0

________________________________________________________

<h3>Simplify the expression using the symmetry of trigonometric function</h3>

cosA=0

________________________________________________________

<h3>Solve the trigonometric equation to find a particular solution</h3>

A=\frac{\pi }{2}\ or\ A=\frac{3 \pi}{2}

________________________________________________________

<h3>Solve the trigonometric equation to find a general solution</h3>

A=\frac{\pi}{2}+2n\pi\ or\ A=\frac{3\pi}{2}+2n\pi,n\in\ Z

________________________________________________________

<h3>Find the union of solution sets</h3>

A=\frac{\pi}{2}+n\pi

________________________________________________________

<h2>A = π/8 + nπ/4 or A = π/2 + nπ, n ∈ Z</h2>

<h3>Find the union of solution sets</h3>

A=\frac{\pi}{8}+\frac{n\pi}{4}\ or\ A=\frac{\pi}{2}+n\pi ,n\in Z

<em>I hope this helps you</em>

<em>:)</em>

5 0
2 years ago
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