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3 years ago

What are the coordinates of the vertices of the triangle under the translation (x, y) mc011-2.jpg (x + 1, y - 4)?

Mathematics

2 answers:

Dima020 [189]3 years ago

8 0

Original
(-1,1) (-4,1) (-4, 4)

 translation (x, y) to (x + 1, y - 4)
(0,-3) (-3,-3) (-3, 0)

answer
(0, -3), (-3, -3), (-3, 0)

luda_lava [24]3 years ago

5 0

it is c :

(0,-3), (-3,-3), (-3,0)

just take the X and Y of the coordinates of the original triangle and substitute it in the given equation (x + 1, y - 4)

try it out yourself it works :)

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For the equation - 3x + 2 = 8 and the inequality

Anarel [89]

Answer:

x= -2; x ≥ 2

Step-by-step explanation:

1. -3x+2=8

Subtract +2 on both sides

-3x+2-2=8-2

-3x=6 Divide by -3 on both sides

-3x/-3=6/-3

x= -2

-2 is the answer for both the equation and inequality.

sorry i was lazy to do the inequality one, but it should be the same answer.

also ignore the fact that theres a line, i could not find the sign without one, o im sorry.

8 0

2 years ago

In a laboratory experiment a student obtained the following values for the mass density of a metal sample: 5.13 g/cc, 4.98 g/cc,

VMariaS [17]

Answer:

5.104%

Step-by-step explanation:

Given:

The accepted value for the mass density = 4.80 g/cc

Observed values:

5.13 g/cc

4.98 g/cc

5.06 g/cc

5.01 g/cc

Total number of observed values = 4

Mean of the observed values = ( 5.13 + 4.98 + 5.06 + 5.01 ) / 4 = 5.045 g/cc

Now,

The percentage error is calculated as:

$\textup{Percent Error}=\frac{\left |\textup{Accepted value-Mean value} \right |}{\left |\textup{Accpeted value} \right |}\times100$ %

on substituting the values, we get

$\textup{Percent Error}=\frac{\left |4.80-5.045\right |}{4.80}\times100$ %

Percentage error = 5.104%

3 0

2 years ago

From a deck of five cards numbered 2, 4, 6, 8, and 10, respectively, a card is drawn at random and replaced. this is done three

Serga [27]

Since the sum of the numbers on the three draws is 12, if we want the card numbered 2 to be drawn exactly two times, the third card can only be numbered 8. In fact, $2+2+8 = 12$ , and there are no other possibilities, unless you consider the various permutations of the terms.

So, we have three favourable cases: we can draw 2,2,8, or 2,8,2, or 8,2,2. This are the only three cases where the card numbered 2 is drawn exactly two times, and the sum of the number on the three draws is 12.

Now, the question is: we have three favourable cases over how many? Well, we have 5 possible outcomes with each draws, and the three draws are identical, because we replace the card we draw every time.

So, we have 5 possible outcomes for the first draw, 5 for the second and 5 for the third. This leads to a total of $5 \times 5 \times 5 = 5^3 = 125$ possible triplets.