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3 years ago

What is 1 name for a square besides a quadrilateral and a square?

Mathematics

1 answer:

telo118 [61]3 years ago

5 0

parallelogram

I hope this helps!

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It takes 3 hours to wash every window on a building.there are 30 windows. How many minutes does it takes to wask each window

Sveta_85 [38]

Answer:6minutes

Step-by-step explanation:

3 hours =180min

180min divided by 30 windows is 6

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The first bus stop on a bus route is 4 miles from school how many yards is the first bus stop from school

Aleksandr [31]

7,040 Yards because there is 1760 yards in a mile.
1760 x 4 = 7,040

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3 years ago

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NISA [10]

Answer:

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Step-by-step explanation:

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3 years ago

The equation of a line that contains the points (4, -3) and (8, -3) is y = -3.

Orlov [11]

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4 years ago

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points C,D, and E are collinear on CE, and CD:DE = 3/5. C is located at (1,8), D is located at (4,5), and E is located at (x,y).

kap26 [50]

Answer:

The point E is located at (9,0)

x=9, y=0

Step-by-step explanation:

we have that

Points C,D, and E are collinear on CE

Point D is between point C and point E

we know that

$CE=CD+DE$ -----> equation A (by addition segment postulate)

$\frac{CD}{DE}=\frac{3}{5}$

$CD=\frac{3}{5}DE$ ------> equation B

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

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

Find the distance CD 

we have

C(1,8), D(4,5)

substitute in the formula

$CD=\sqrt{(5-8)^{2}+(4-1)^{2}}$

$CD=\sqrt{(-3)^{2}+(3)^{2}}$

$CD=\sqrt{18}\ units$

Find the distance DE

substitute the value of CD in the equation B and solve for DE

$\sqrt{18}=\frac{3}{5}DE$

$DE=\frac{5\sqrt{18}}{3}\ units$

Find the distance CE

$CE=CD+DE$

we have

$DE=\frac{5\sqrt{18}}{3}\ units$

$CD=\sqrt{18}\ units$

substitute the values in the equation A

$CE=\sqrt{18}+\frac{5\sqrt{18}}{3}$

$CE=\frac{8\sqrt{18}}{3}$

Applying the formula of distance CE

we have

$CE=\frac{8\sqrt{18}}{3}$

C(1,8), E(x,y)

substitute in the formula of distance

$\frac{8\sqrt{18}}{3}=\sqrt{(y-8)^{2}+(x-1)^{2}}$

squared both sides

$128=(y-8)^{2}+(x-1)^{2}$ -----> equation C

Applying the formula of distance DE

we have

$DE=\frac{5\sqrt{18}}{3}\ units$

D(4,5), E(x,y)

substitute in the formula of distance

$\frac{5\sqrt{18}}{3}=\sqrt{(y-5)^{2}+(x-4)^{2}}$

squared both sides

$50=(y-5)^{2}+(x-4)^{2}$ -----> equation D

we have the system

$128=(y-8)^{2}+(x-1)^{2}$ -----> equation C

$50=(y-5)^{2}+(x-4)^{2}$ -----> equation D

Solve the system by graphing

The intersection point both graphs is the solution of the system

The solution is the point (9,0)

therefore

The point E is located at (9,0)

see the attached figure to better understand the problem

6 0

3 years ago