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likoan [24]
4 years ago
10

Helppp ASAP pls thank you

Mathematics
1 answer:
lukranit [14]4 years ago
7 0
C. The number of different colors on the page
You might be interested in
Find the midsegment of triangle ABC so that it is parallel to side BC and label it as segment EG on the graph.
Lelechka [254]

Answer:

Coordinate of E:(-2,3) Coordinate of G: (2.5,1.5)

Parallel: slopes both = -1/3

1/2 segment: EG length= 3 BC length=6: This is the only one I'm not sure about.

I will show the work for both the problems down below.

Step-by-step explanation:

Finding coordinates:

1+(-5)/2=-2  5+1/2=3

(-2,3)

1+4/2= 2.5   5+(-2)/2= 1.5

(2.5,1.5)

Show work verifying the EG and BC are parallel:

You can do this by finding the slope

Slop formula (y2-y1)/(x2-x1)

(1.5-3)/(2.5-(-2))

Then simplify

-1.5/2/5= -1/3 which i your slope. You use the same formula and follow thee same steps for the points of BC and both will come up as -1/4 meaning they are parallel.

Show work verifying that EG and BC are parallel:

Use distance formula to find how long each line is

√(x2-x1)^2+(y2-y1)^2

√(2.5-(-2)^2)+(1.5-3)^2

Simplify

√4.5+(1.5)=3

Do the same thing with BC and you will get 6






7 0
3 years ago
AB and CD are straight lines. Find AOD.​
Anon25 [30]

Answer:

106°

Step-by-step explanation:

AOD = COB (because of vertical angles)

COB = COF + FOB

COF = 90

FOB = 16

COB = 90 + 16 = 106

AOD = 106

Hope this helps :)

Have a nice day!

3 0
3 years ago
Find all points having an x-coordinate of 5 whose distance from the point (2, 6) is 5
galina1969 [7]

Answer:

(5,2) and (5,10)

Step-by-step explanation:

we know that

the formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

we have

points\ (2,6),(5,y)\\d=5

substitute in the formula

5=\sqrt{(y-6)^{2}+(5-2)^{2}}

solve for y

5=\sqrt{(y-6)^{2}+(3)^{2}}

5=\sqrt{(y-6)^{2}+9}

square both sides

25=(y-6)^{2}+9

(y-6)^{2}=25-9

(y-6)^{2}=16

square root both sides

y-6=\pm4

y=6\pm4

y=6+4=10

y=6-4=2

therefore

The points are

(5,2) and (5,10)

6 0
3 years ago
Estimate the size of a crowd walking in a charity fundraising march that occupies a rectangular space with dimensions of 10 feet
jolli1 [7]

Answer:

4000 is size of a crowd walking in a charity fundraising March.

Step-by-step explanation:

Given:

Dimensions of rectangular space 10 feet by 12000 feet

So we can say that,

Length = 1200 ft

Width = 10 ft

Now we will calculate the area of rectangular space which is given by

Hence Area will be = length\times width= 10\ ft \times 1200 \ ft= 12000 \ ft^2

Now we know that 40 people occupy a rectangle measuring 10 feet by 12 feet.

Length = 12 ft

Width = 10 ft

Area = length\times width= 10\ ft \times 12 \ ft= 120 \ ft^2

It says that 40 people are occupied in 120 ft^2

So how many people will be there in 12000 ft^2

By using unitary method we get,

Number of people = \frac{40\times 12000 \ ft^2}{120 \ ft^2} = 4000 \ peoples

Size of a crowd walking in a charity fundraising March is 4000.

8 0
4 years ago
Read 2 more answers
Plz help with 1 through 6 show ur work
NeX [460]
1. 5×10×7= 350
2. 12×6×8=576
3. 4.2×3.5×2=29.4
4. 1.1×1.5×2.6=4.29
5.4×3×5= 60
6. 4.1×2.6×5.1=47.97
5 0
3 years ago
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