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Crank
2 years ago
8

Someone please help me with my mathematics work!

Mathematics
1 answer:
harina [27]2 years ago
3 0

Answer:

108 Ft^2

Step-by-step explanation:

So the formula for finding the area of a trapezoid is Area= 1/2(a+b)*h

So in this equation, your a=6ft and your b=18ft and your h=9ft

Add the two parallel lines (a and b) and then divide that product by 2. Then multiply that product by 9.

This should give you a total of 108 ft^2 (squared)

Hope this helps!

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A candle has been burning for 20 min and is now 25 cm tall. In an hour it will be 10 cm tall. Which equation models the height y
stiv31 [10]

Answer:

(y - 25) = - 0.25(x - 20)

Step-by-step explanation:

Given that :

Height of candle after burning for 20 minutes = 25 cm

Height after burning for 1 hr (60 minutes) = 10 cm

Height (y) in cm of candle x minutes after being lit:

Using the equation :

(y - y1) = m(x - x1)

m = (change in y / change in x)

Change in height within 60 minutes :

Height at 20 minutes = 25cm

Height after an hour = 10

Change in height per hour = (25 - 10) = 15cm

Hence, m = change in height per minute

15cm / 60 = 0.25cm ( - 0.25) (decrease in height)

y1 = 25 ; x1 = 20

(y - y1) = m(x - x1)

(y - 25) = - 0.25(x - 20)

4 0
3 years ago
The glee club has a set of 48 classical cds and a set of 42 show tune cds. each set can be divided equally among the club member
Hoochie [10]
We have to find the GCF (Greatest Common Factor) of 48 and 42 here... 

Factors of 48:
1,2,3,4,6,8,12,16,24,48

Factors of 42: 
1,2,3,6,7,14,21,42

The largest common factor of these two numbers is 6. 


So, there can be a maximum of 6 members in the club.. 

Hope this helped!
6 0
4 years ago
Read 2 more answers
Find the Slope of lines passing through the points: (4,5) and (6,7)​
kupik [55]

the answer is 0.6 you gotta divide and multiply

4 0
3 years ago
Find all polar coordinates of point p where p = (6, -pi/5)
Lelu [443]
Given a point in coordinates form (r,\theta), one can compute the cartesian form like this:
(r\cos\theta,r\sin\theta )
We have:
r=6,\theta=-\dfrac{\pi}{5}
We get the cartesian form then:
\left(6\cos-\dfrac{\pi}{5},6\sin-\dfrac{\pi}{5}\right)
4 0
3 years ago
The Cartesian coordinates of a point are given. (a) (−5, 5) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0
Alex73 [517]

Answer:

a)

(i) The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) The coordinates of the point in polar form is (-5√2 , 3π/4)

b)

(i) The coordinates of the point in polar form is (6 , π/3)

(ii) The coordinates of the point in polar form is (-6 , 4π/3)

Step-by-step explanation:

* Lets study the meaning of polar form

- To convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):

1. r = √( x2 + y2 )

2. θ = tan^-1 (y/x)

- In Cartesian coordinates there is exactly one set of coordinates for any

 given point

- In polar coordinates there is literally an infinite number of coordinates

 for a given point

- Example:

- The following four points are all coordinates for the same point.

# (5 , π/3) ⇒ 1st quadrant

# (5 , −5π/3) ⇒ 4th quadrant

# (−5 , 4π/3) ⇒ 3rd quadrant

# (−5 , −2π/3) ⇒ 2nd quadrant

- So we can find the points in polar form by using these rules:

 [r , θ + 2πn] , [−r , θ + (2n + 1) π] , where n is any integer

 (more than 1 turn)

* Lets solve the problem

(a)

∵ The point in the Cartesian plane is (-5 , 5)

∵ r = √x² + y²

∴ r = √[(5)² + (-5)²] = √[25 + 25] = √50 = ±5√2

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (5/-5) = tan^-1 (-1)

- Tan is negative in the second and fourth quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = 2π - tan^-1(1) ⇒ in fourth quadrant r > 0

∴ Ф = 2π - π/4 = 7π/4

OR

∴ Ф = π - tan^-1(1) ⇒ in second quadrant r < 0

∴ Ф = π - π/4 = 3π/4

(i) ∵ r > 0

∴ r = 5√2

∴ Ф = 7π/4 ⇒ 4th quadrant

∴ The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) r < 0

∴ r = -5√2

∵ Ф = 3π/4 ⇒ 2nd quadrant

∴ The coordinates of the point in polar form is (-5√2 , 3π/4)

(b)

∵ The point in the Cartesian plane is (3 , 3√3)

∵ r = √x² + y²

∴ r = √[(3)² + (3√3)²] = √[9 + 27] = √36 = ±6

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (3√3/3) = tan^-1 (√3)

- Tan is positive in the first and third quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = tan^-1 (√3) ⇒ in first quadrant r > 0

∴ Ф = π/3

OR

∴ Ф = π + tan^-1 (√3) ⇒ in third quadrant r < 0

∴ Ф = π + π/3 = 4π/3

(i) ∵ r > 0

∴ r = 6

∴ Ф = π/3 ⇒ 1st quadrant

∴ The coordinates of the point in polar form is (6 , π/3)

(ii) r < 0

∴ r = -6

∵ Ф = 4π/3 ⇒ 3rd quadrant

∴ The coordinates of the point in polar form is (-6 , 4π/3)

6 0
4 years ago
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