Please right answer only immediately

Vlada [557]

Fiidjdjdjdsjjdjdjdje

7 0

3 years ago

Mulder and Scully are driving to the same town. Mulder leaves the office at 9:30a.m. averaging 57mph. Scully leaves at 10:00a

koban [17]

Answer:

Scully will catch up with Mulder by 7:30 PM.

Step-by-step explanation:

Consider the provided information.

Mulder leaves the office at 9:30a.m. averaging 57mph.

Scully leaves at 10:00a.m., following the same path and averaging 60 mph.

Distance covered by Mulder between 9:30 AM and 10:00 AM:

$\frac{57}{2}= 28.5$ miles

Let t is the time taken by Scully to catch up with Mulder.

After leaving office Scully needs to cover extra distance of 28.5 miles because Mulder leaves the office earlier.

Therefore total distance cover by them is:

$28.5+57t=60t$

$28.5=60t-57t$

$28.5=3t\\t=9.5$

Hence, it would take 9 hours 30 minutes to catch up Mulder.

10 am + 9 hours 30 minutes = 7:30 pm

Therefore, Scully will catch up with Mulder by 7:30 PM.

6 0

3 years ago

How was it like living in the 1900s-1800s

WINSTONCH [101]

Answer:

OML

Step-by-step explanation:

How would anyone know this

3 0

3 years ago

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

Here is free points FOR BEINGS NERDS

Maurinko [17]

Answer:

thank u

Step-by-step explanation:

3 0

3 years ago

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