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NeTakaya
3 years ago
6

Triangle ABC is to be translated right 3 units, down 5 units. What are the vertices of the translated triangle? A. A'(0, 9), B'(

1, 3), and C'(7, 6) B. A'(0, –1), B'(1, –7), and C'(7, –4) C. A'(–6, –1), B'(–5, –7), and C'(1, –4) D. A'(–6, 9), B'(–5, 3), and C'(1, 6)

Mathematics
1 answer:
jeka57 [31]3 years ago
6 0
It is A. Because the point a would be on the -1 and the b would be on the -7 and the c would be on the -4
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1 is 25% of what numder
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Answer:

4, it's obvious think of it as a quarter.

Step-by-step explanation:

1 = 25%

2 = 50%

3 = 75%

4 = 100%

5 0
3 years ago
Please dont ignore, Need help!!! Use the law of sines/cosines to find..
Ket [755]

Answer:

16. Angle C is approximately 13.0 degrees.

17. The length of segment BC is approximately 45.0.

18. Angle B is approximately 26.0 degrees.

15. The length of segment DF "e" is approximately 12.9.

Step-by-step explanation:

<h3>16</h3>

By the law of sine, the sine of interior angles of a triangle are proportional to the length of the side opposite to that angle.

For triangle ABC:

  • \sin{A} = \sin{103\textdegree{}},
  • The opposite side of angle A a = BC = 26,
  • The angle C is to be found, and
  • The length of the side opposite to angle C c = AB = 6.

\displaystyle \frac{\sin{C}}{\sin{A}} = \frac{c}{a}.

\displaystyle \sin{C} = \frac{c}{a}\cdot \sin{A} = \frac{6}{26}\times \sin{103\textdegree}.

\displaystyle C = \sin^{-1}{(\sin{C}}) = \sin^{-1}{\left(\frac{c}{a}\cdot \sin{A}\right)} = \sin^{-1}{\left(\frac{6}{26}\times \sin{103\textdegree}}\right)} = 13.0\textdegree{}.

Note that the inverse sine function here \sin^{-1}() is also known as arcsin.

<h3>17</h3>

By the law of cosine,

c^{2} = a^{2} + b^{2} - 2\;a\cdot b\cdot \cos{C},

where

  • a, b, and c are the lengths of sides of triangle ABC, and
  • \cos{C} is the cosine of angle C.

For triangle ABC:

  • b = 21,
  • c = 30,
  • The length of a (segment BC) is to be found, and
  • The cosine of angle A is \cos{123\textdegree}.

Therefore, replace C in the equation with A, and the law of cosine will become:

a^{2} = b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}.

\displaystyle \begin{aligned}a &= \sqrt{b^{2} + c^{2} - 2\;b\cdot c\cdot \cos{A}}\\&=\sqrt{21^{2} + 30^{2} - 2\times 21\times 30 \times \cos{123\textdegree}}\\&=45.0 \end{aligned}.

<h3>18</h3>

For triangle ABC:

  • a = 14,
  • b = 9,
  • c = 6, and
  • Angle B is to be found.

Start by finding the cosine of angle B. Apply the law of cosine.

b^{2} = a^{2} + c^{2} - 2\;a\cdot c\cdot \cos{B}.

\displaystyle \cos{B} = \frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}.

\displaystyle B = \cos^{-1}{\left(\frac{a^{2} + c^{2} - b^{2}}{2\;a\cdot c}\right)} = \cos^{-1}{\left(\frac{14^{2} + 6^{2} - 9^{2}}{2\times 14\times 6}\right)} = 26.0\textdegree.

<h3>15</h3>

For triangle DEF:

  • The length of segment DF is to be found,
  • The length of segment EF is 9,
  • The sine of angle E is \sin{64\textdegree}}, and
  • The sine of angle D is \sin{39\textdegree}.

Apply the law of sine:

\displaystyle \frac{DF}{EF} = \frac{\sin{E}}{\sin{D}}

\displaystyle DF = \frac{\sin{E}}{\sin{D}}\cdot EF = \frac{\sin{64\textdegree}}{39\textdegree} \times 9 = 12.9.

7 0
3 years ago
Determine whether the following events are mutually exclusive:
Harman [31]

Answer:

non mutually exclusive event

Step-by-step explanation:

im not sure on this tho

3 0
3 years ago
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Hi i would like sum help :)) please help for a brainlist !!!!!!!!!!!!!
Sergio [31]
A. all real numbers for both

Explanation:
Domain is the x-values. Since the graph goes forever horizontally, any x-value is on the graph.
Range is the same but with y-values. If the graph went on further than what the picture shows, then it also would show that every y value is on the graph
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3 years ago
Find the gradient of the line segment between the points (0,2) and (-2,10).​
Tasya [4]

Answer:

-4

Step-by-step explanation:

Find the gradient of the line segment between the points (0,2) and (-2,10).​

Given data

x1= 0

x2= -2

y1= 2

y2=10

The expression for the gradient is given as

M= y2-y1/x2-x1

substitute

M= 10-2/-2-0

M= 8/-2

m= -4

Hence the gradient is -4

5 0
3 years ago
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