Ask question

Ask question

All categories

3 years ago

11

Band students at Lowell High School sell candy every year as a fundraiser. Last year, they sold 90 boxes of truffles and 77 boxe

s of peanut brittle, raising a total of $835. This year, they sold 62 boxes of truffles and 66 boxes of peanut brittle, from which they raised $640. How much does the band earn from each item?

1 answer:

azamat3 years ago

3 0

The band earns 5 dollars from each item.

You might be interested in

Hypoteneous base and what is the other

Yuliya22 [10]

Answer:

Just the hypotenuse and the 2 legs

7 0

3 years ago

Read 2 more answers

Domain and range???!!!!!!

elena55 [62]

Answer:

domain={1,-3}

range={-5,1,2,4}

6 0

3 years ago

Please answer all please

Firdavs [7]

Answer:

Step-by-step explanation:

The first parabola has vertex (-1, 0) and y-intercept (0, 1).

We plug these values into the given vertex form equation of a parabola:

y - k = a(x - h)^2 becomes

y - 0 = a(x + 1)^2

Next, we subst. the coordinates of the y-intercept (0, 1) into the above, obtaining:

1 = a(0 + 1)^2, and from this we know that a = 1. Thus, the equation of the first parabola is

y = (x + 1)^2

Second parabola: We follow essentially the same approach. Identify the vertex and the two horizontal intercepts. They are:

vertex: (1, 4)

x-intercepts: (-1, 0) and (3, 0)

Subbing these values into y - k = a(x - h)^2, we obtain:

0 - 4 = a(3 - 1)^2, or

-4 = a(2)². This yields a = -1.

Then the desired equation of the parabola is

y - 4 = -(x - 1)^2

7 0

3 years ago

LM = JK then JK = LM<br> What property is this?

konstantin123 [22]

This example is the Symmetric Property.

I hope this helps you!
xo, Leafling

5 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