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oksian1 [2.3K]
3 years ago
6

Please help me

Mathematics
1 answer:
Sauron [17]3 years ago
3 0

Answer:

b

Step-by-step explanation:

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At fairvie middle school 75 band members need to raise a total of 8250 for a trip. So far they have raised 3120. How much money
galina1969 [7]
$8250 - $3120 = $5130

$5130/75 = $68.40

Each band member must raise $68.40
8 0
3 years ago
The lassos high school senior class raised funds for an end of the year cruise getaway. The city gave them a special package for
kirill115 [55]

Answer:

Step-by-step explanation:

x = total number of students

1500 + 367x = 16,600

367x = 16,600 - 1500

367x = 15100

x = 15100 / 367

x = 41.14.......rounds to 41 students <==

this only works if the port fees and taxes that the city gave them are included in the money the class raised

5 0
3 years ago
I need help i need to get this right please i begggggg you<br>i need help ASAP
Flauer [41]
Please say brainliest.

The answer is yes.

Plug 4 in for the value of n so you have 4+8 is less than it equal to 13. And since 12 is less than 13 it is true, and a solution to the inequality.
4 0
3 years ago
I NEED HELP !!!!!!!!!!!
madreJ [45]
The slope is found by rise over run so it’s 8/4 and that equals 2 so it’s 2x and the second number is the y intercept which is .5 so the overall answer is
y= 2x + 0.5
5 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
Arada [10]

Given:

Scale factor s=\dfrac{1}{3}

Center of dilation = (4,2)

To find:

The coordinates of the points C' and A.

Solution:

We know that, if a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

The scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Suppose the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using rule (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Hence, the coordinates of Point C' are C'(2,5).

Let us assume that point A is A(m,n).

Using rule (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

(m-4)=3(4)

m=12+4

m=16

And,

\dfrac{1}{3}(n-2)+2=4

\dfrac{1}{3}(n-2)=4-2

(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

4 0
2 years ago
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