A trip from nyc to seattle is 2;852.1 miles . A family want to make the drive in 10 days

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

Suppose you needed $ 22 comma 000$22,000 to maintain a particular standard of living in 19781978. How much would you have needed

wlad13 [49]

Sorry i dont know this is kind of tricky

5 0

3 years ago

Evaluate the sum of the finite geometric series. 1+2+4+8+...+128

Wittaler [7]

S = a*((1 - r^(n+1))/(1-r)) S = 1*((1 - 2^(7+1))/(1-2)) S = 1*((1 - 2^8)/(1-2)) S = 1*((1 - 256)/(1-2)) S = 1*((-255)/(-1)) S = 255 So 1+2+4+8+...+128 = 255

7 0

3 years ago

Can you help me? sin 35 (degrees) Thanks!

Nata [24]

OK, here's a *very* approximate "creative" method:

sin(a+b)=sin(a)cos(b) + cos(a)sin(b) 

choose a=30deg and b=5deg 
- sin(a)=0.5 (exact)
- b is a small number so we say sin(b)~b (in radians). We can do a long division for this: 3.14/36 = 0.087 radians
- similarly, for b sufficiently small, cos(b)~1
- now we only need cos(a) = sqrt(3)/2. We can approximately calculate sqrt(3) as 1.73, (by trial-and-error or better, by using the Babylonian method; so sqrt(3)/2 = 0.865 

Finally, sin(30 + 5) = (0.5 x 1) + (0.087 x 0.865) = 0.575 (calculating more digits is of no use because the sin(b)=b and cos(b)=1 are quite inaccurate approximations). 

So that wasn't all too much effort, and we're close to the correct answer (0.574)

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

8 0

3 years ago

7) One track coach wants his athletes to race 5 miles around a track to measure how fast

Lady bird [3.3K]

Five laps? I don’t know if you meant to put that the track is a mile around, but if the track is a half mile around, an athlete would have to run 10 laps around the track to run 5 miles.
X=5/y, so x would be how many laps the athletes run, and y would be the length of the track. Hope this helps

5 0

3 years ago