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Ilya [14]
2 years ago
14

If ⅔ part of a land cost Rs 45000 find the cost of ⅘ part of land​

Mathematics
1 answer:
Vika [28.1K]2 years ago
6 0

Answer:

GG from Shahs I am not a good time to meet you and your company and the other side of the year and the other hand the year and a good time and money by simply beautiful day in the future of our lives to the other hand I

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Yesterday morning at 8 a.m. the temperature was -14˚F. At noon it was 10 degrees warmer. The temperature increase between noon a
OLEGan [10]
The temperature was -14°F at 8 a.m.

At noon is was 10 degrees warmer - the temperature increase was 10°F
-14°F + 10°F = -4°F

The tempertaure increase between noon and 4 p.m. was twice the previous increase in temperature, which was 10°F.
10°F x 2 = 20°F

We want to know the temperature at 4 p.m. so we just add it up to the previous result.
-4°F + 20°F = 16°F

16°F is the correct answer
4 0
3 years ago
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You want to purchase a ticket to the movies. A ticket costs at least $12. You have $3. How much more money do you need to purcha
BaLLatris [955]

Answer:

I think the answer is$12 - $3 = $9

the amount u need is $9 to have the complete amount u need to purchase the ticket

6 0
2 years ago
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كيفية حساب cos a و tan a بحيث sin a يساوي 1على 2
Paraphin [41]

Answer:

uh i dont know

Step-by-step explanation:

8 1
2 years ago
How to find equation hyperbola with asymptotes, center and transverse?
lapo4ka [179]
The equation for a hyperbola is (x-h)/a - (y-k)/b = 1
Or (y-k)/a - (x-h)/b = 1
h represents the x value of the coordinate
k value represents the y value of the coordinate 
together they represent a point, which is the center
So (h,k) is (x,y)

The asymptote is y-k = +/- b/a (x-h)
The transverse is the line that goes through the hyperbola. 
6 0
3 years ago
The center of a hyperbola is located at the origin. One focus is located at (−50, 0) and its associated directrix is represented
leva [86]

The equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

The center of a hyperbola is located at the origin that means at (0, 0) and one of the focus is at (-50, 0)

As both center and the focus are lying on the x-axis, so the hyperbola is a horizontal hyperbola and the standard equation of horizontal hyperbola when center is at origin: \frac{x^{2}}{a^{2}}  - \frac{y^{2}}{b^{2}}    = 1

The distance from center to focus is 'c' and here focus is at (-50,0)

So, c= 50

Now if the distance from center to the directrix line is 'd', then

d= \frac{a^{2}}{c}

Here the directrix line is given as : x= 2304/50

Thus, \frac{a^{2}}{c}  = \frac{2304}{50}

⇒ \frac{a^{2}}{50}  = \frac{2304}{50}

⇒ a² = 2304

⇒ a = √2304 = 48

For hyperbola, b² = c² - a²

⇒ b² = 50² - 48² (By plugging c=50 and a = 48)

⇒ b² = 2500 - 2304

⇒ b² = 196

⇒ b = √196 = 14

So, the equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

5 0
3 years ago
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