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kotykmax [81]
3 years ago
11

On a coordinate plane, triangle A B C is shifted 4 units to the right and 2 units up to form triangle A prime B prime C prime. I

dentify the rule for the transformation shown in the figure that translates triangle ABC into triangle A'B'C'. (x, y) Right-arrow (x – 4, y – 2) (x, y) Right-arrow (x + 4, y – 2) (x, y) Right-arrow (x – 4, y + 2) (x, y) Right-arrow (x + 4, y + 2)
Mathematics
1 answer:
zheka24 [161]3 years ago
3 0

Answer:

(x, y) ⟶ (x + 4, y + 2)  

Step-by-step explanation:

If you shift a point four units to the right, x ⟶ x + 4.

If you shift it up two units, y ⟶ y + 2.

If you combine the two shifts into one transformation, the rule becomes

(x, y) ⟶ (x + 4, y + 2)

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What is the slope of a line that is parallel to the graph of the equation y = 10x - 20?
8_murik_8 [283]

10

The slope of all parallel lines is equal. 10 is the slope in the equation y = 10x - 20, so the slope of a line parallel to it has to be 10 also.

4 0
3 years ago
Please help! Help Will Give 100 PTS
allochka39001 [22]

Answer:

  x = 23; not extraneous

Step-by-step explanation:

A solution is extraneous if it does not satisfy the original equation. Extraneous solutions can sometimes be introduced in the process of solving radical and rational function equations.

<h3>Solution</h3>

Squaring both sides of the given equation, we get ...

  √(3x +12) = 9

  3x +12 = 81 . . . . . . square both sides

  x +4 = 27 . . . . . . . divide by 3

  x = 23 . . . . . . . . . . subtract 4

<h3>Check</h3>

There is only one solution, and it satisfies the equation:

  √(3×23 +12) = √81 = 9

The solution x = 23 is not extraneous.

7 0
2 years ago
If (cos)x= 1/2 what is sin(x) and tan(x)? Explain your steps in complete sentences.
Natasha_Volkova [10]
The correct answers are:
1) sin(x) = \frac{ \sqrt{3} }{2}
2) tan(x) = \sqrt{3}

Explanation:
Given:
cos(x) =  \frac{1}{2}

Step 1:
Since, according to the Trigonometric identity:
sin^2(x) + cos^2(x) = 1 -- (1)

Step 2:
Plug in the value of cos(x) in equation (1):
sin^2(x) + ( \frac{1}{2} )^2 = 1 \\ sin^2(x) + \frac{1}{4} = 1 \\ sin^2(x)  =  \frac{3}{4}

Step 3:
Take square-root on both sides:
\sqrt{sin^2(x)} =  \sqrt{\frac{3}{4}}

sin(x) = \frac{ \sqrt{3} }{2}

Now to find the tan(x), we would use the following formula:

tan(x) = \frac{sin(x)}{cos(x)} --- (2)

Plug in the values of sin(x) and cos(x) in equation (2):
tan(x) = \frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} }

Hence tan(x) = \sqrt{3}
6 0
3 years ago
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SashulF [63]
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DerKrebs [107]

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