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neonofarm [45]
3 years ago
13

in the 2010 new york city marathon 42429 people finished the race and received a medal before the race the medals had to be orde

red if you were the person in charge of ordering the medals and estimated how many to order by rounding would u have ordered enough medals explain your thinking
Mathematics
1 answer:
Rufina [12.5K]3 years ago
5 0
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yuradex [85]
You can find the volume by multiplying the Length X width X height.
6 0
3 years ago
I need help pleaseeeee!!
PilotLPTM [1.2K]

Answer:

option 4

Step-by-step explanation:

fog(x)=(x-5)²+4

=x²-10x+25+4

fog(x)=x²-10x+29

8 0
3 years ago
Corey spent 20% of his savings on a printer at Louie's ElectronisHow much did Corey have in his savings account before he bought
Crank
(printer cost) = 0.20 * (savings)
(printer cost)/0.20 = (savings)

savings = 5*(printer cost)

Whatever the cost of the printer was (information not supplied here), Corey's savings was 5 times that amount.
5 0
3 years ago
10th grade level LT1
Slav-nsk [51]

Answer:

<h2>? = 3.</h2>

Hope this will help you.

5 0
3 years ago
The green triangle is a dilation of the red triangle with a scale factor of s=1/3 and the center of dilation is at the point (4,
klasskru [66]

Given:

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

Red is original figure and green is dilated figure.

To find:

The coordinates of point C' and point A.

Solution:

Rule of dilation: 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)

According to the given information, 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)

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

Using (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)

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

We assumed that point A is A(m,n).

Using (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).

5 0
3 years ago
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