I REALLY NEED HELP ITS FOR A MAJOR GRADE TEST<br> CORRECT = BRAINLST

The height in meters of a toy rocket at t seconds can be found by the

OlgaM077 [116]

Answer: 81.2

Step-by-step explanation:

h(t) = -5t² + 40t + 1.2

h(4) = -5(4)² + 40(4) + 1.2 We plug 4 seconds as t, and solve h(4)

h(4) = -80 + 160 + 1.2

h(4) = 80 + 1.2

h(4) = 81.2

4 0

3 years ago

Molli purchased 8 apples for $2.96 while looking at her reciept molli divided $2.96 by 8. Why might molli perform this calculati

Basile [38]

B

Let’s put this into a simple equation

8 apples = 2.96

We can substitute apples for x

8x=2.96

She divided by 8

X= 2.96/8

Remember x represents apples so only 1 x represents 1 apple

So that means the answer is B.

Hope that helped!

4 0

3 years ago

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Depenendent Event

DaniilM [7]

Answer:

.2

Step-by-step explanation:

first you add all the pencils up which is 12 next take the probably of taking out one of each pencil .4 for green .2 for black and .6 for yellow next you divide the probably for pulling 1 green bye 2 since your pulling out 2

8 0

3 years ago

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Geometry help ixl asap pls !

QveST [7]

Answer:

10 i think

Step-by-step explanation:

3 0

3 years ago

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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

I REALLY NEED HELP ITS FOR A MAJOR GRADE TEST CORRECT = BRAINLST