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valina [46]
3 years ago
15

Given that A'B'C' is the image of ABC under a reflection, enter that equation of the line of reflection.

Mathematics
1 answer:
PolarNik [594]3 years ago
7 0

Answer:

sooooo

Step-by-step explanation:

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The sum of two numbers is 21. The difference between them is 5. What are the two numbers?
Nata [24]

Answer: 13 and 8


Step-by-step explanation:

13+8=21

13-8=5

All I did was trial and error. Plus, it helps to set up an equation : x+x=21, x-x=5

5 0
4 years ago
In ABC, if the lengths of sides a and c are 8 centimeters and 16 centimeters, respectively, and the measure of C is 35 degrees,
Simora [160]

Answer:

∠ A ≈ 16.67°

Step-by-step explanation:

Using the Sine rule in Δ ABC

\frac{a}{sinA} = \frac{c}{sinC} , substitute values

\frac{8}{sinA} = \frac{16}{sin35} ( cross- multiply )

16 sinA = 8 sin35° ( divide both sides by 16 )

sin A = \frac{8sin35}{16} , thus

∠ A = sin^{-1} ( \frac{8sin35}{16} ) ≈ 16.67° ( to 2 dec. places )

8 0
3 years ago
(2.01)A polygon is shown on the graph:
romanna [79]

3 units down would be subtracting 3 from the  Y values and 4 units left would be subtracting 4 from the X values

Point A is at (2,-1)

The new point would be at (2-4, -1 -3) = (-2,-4)

Point B is at (1,-4)

The new point would be at (1-4, -4 -3) = (-3,-7)

Point C is at (3,-5)

The new point would be at (3-4, -5-3) = (-1,-8)

Point D is at (5,-3)

The new point would be (5-4, -3-3) = (1,-6)

So the new location would be:

A' (-2,-4)

B' (-3,-7)

C' (-1,-8)

D' (1,-6)

8 0
3 years ago
Read 2 more answers
3 cards are drawn from a standard deck without replacement. What is the probability that at least one of the cards drawn is a re
Rudik [331]

Answer: \dfrac{15}{17}

Step-by-step explanation:

Total number of cards in a deck = 52

Number of red cards = 26

Number of cards not red =

Number of ways to draw not red cards = ^{26}C_3

Total ways to draw 3 cards = ^{52}C_3

The probability that none of three cards are red = \dfrac{^{26}C_3}{^{52}C_3}

=\dfrac{\dfrac{26!}{3!(26-3)!}}{\dfrac{52!}{3!(52-3)!}}  [∵ ^nC_r=\dfrac{n!}{r!(n-r)!}]

=\dfrac{\dfrac{26\times25\times24\times23!}{(23)!}}{\dfrac{52\times51\times50\times49!}{3!(49)!}}=\dfrac{2}{17}

Now , the probability that at least one of the cards drawn is a red card = 1- Probability that none cards are red

=1-\dfrac{2}{17}=\dfrac{17-2}{17}=\dfrac{15}{17}

Hence, the required probability = \dfrac{15}{17}

6 0
3 years ago
Find the coordinates of the point which divides the join of (-1,7) and (4.-3) in the ratio 2:3 ??
d1i1m1o1n [39]
<h2><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u></h2>

Find the coordinates of the point which divides the join of (-1,7) and (4,-3) in the ratio 2:3 ?

<h2><u>Solution</u>:-</h2>

Let the given points be A(-1,7) and B(4,-3)

Now,

Let the point be P(x, y) which divides AB in the ratio 2:3

Here,

<h3>{ \bf  x \: = \frac{m_1 x_2 \: + \: m_1 x_1}{m_1 \: + \: m_2}}</h3>

Where,

m_1 = 2 , m_2 = 3

x_1 = -1 , x_2 = 4

\therefore Putting values we get,

x = \frac{2 \: × \: 4 \: + \: 3 \: × \: -1}{2 \: + \: 3}

x = \frac{8 \: - \: 3}{5}

x = \frac{5}{5}

x = 1

Now,

Finding y

<h3>{ \bf  y \: = \frac{m_1 y_2 \: + \: m_2 y_1}{m_1 \: + \: m_2}}</h3>

Where,

m_1 = 2 , m_2 = 3

x_1 = 7 , x_2 = -3

\therefore Putting values we get,

y = \frac{2 \: × \: -3 \: + \: 3 \: × \: 7}{2 \: + \: 3}

y = \frac{-6 \: + \: 21}{5}

y = \frac{15}{5}

y = 3

Hence x = 1, y = 3

So, the required point is P(x, y)

= P(1, 3)

<h3>The coordinates of the point is P(1, 3). [Answer]</h3>

_______________________________________

<u>N</u><u>o</u><u>t</u><u>e</u>:- Refer the attachment.

_______________________________________

5 0
3 years ago
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