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3 years ago

I need help plez and thanks

Mathematics

2 answers:

erastova [34]3 years ago

8 0

It will take 3 hours.

slamgirl [31]3 years ago

4 0

I got 3 hours hope this helps

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Write an equation of the line below.

statuscvo [17]

Answer:

y = -4x - 4

Step-by-step explanation:

Rise/Run

-8/2 = -4

Y-intercept:

-4

8 0

3 years ago

All right angles are _____ degrees

babymother [125]

I think the answer is 90 degrees

7 0

3 years ago

Osvoldo has a goal of getting at least 30% of his grams of carbohydrates each day from whole grains. Today, he ate 220 grams of

saw5 [17]

Answer:

Osvoldo did not meet his goal

Step-by-step explanation:

To know the percentage of carbohydrates that Osvoldo ate from grains with respect to the total, we simply have to divide the amount of carbohydrates he got with the grains by the total

55 / 220 = 0.25

then to express it as a percentage this number has to be multiplied by 100

0.25 * 100 = 25%

Osvoldo wanted to reach 30% and only got 25%

25% < 30%

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3 years ago

331 students went on a field trip. Six buses were filled and 7 students traveled in cars. How many students were in each bus?

Butoxors [25]

<span>54 students travelled in each bus</span>

5 0

3 years ago

Read 2 more answers

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

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