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vladimir2022 [97]
2 years ago
15

Rajan bought a book for Rs. 180 and sold it to Sajan at a profit of 20% .Sajan sold that book to Nirajan at a loss of 20%. At wh

at price should Nirajan sell the book to receive 5% profit.
( no link pls)​
Mathematics
1 answer:
SIZIF [17.4K]2 years ago
3 0

Answer:

226.8

Step-by-step explanation:

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a washer and a dryer cost $640 combined. the washer cost $60 less than the dryer . what is the cost of the dryer?
Delvig [45]

Answer:

380

Step-by-step explanation:

640/2=320

320-60=260

640-260=380

5 0
4 years ago
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jessica has a jewelry box with a base of 18 square inches. the height is 7 inches. what is the volume of her jewelry box?
ozzi
Assuming it is a right rectangular prism, the volume is 126 sq. in.

V = area of base * height = 18 * 7 = 126
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3 years ago
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PLEASE HELP MEEEEe<br> jhchnthnht
Mars2501 [29]

The equation of the perpendicular bisector of BC with B(-2, 1), and C(4, 2) is y = 7.6 - 6•x

<h3>Which method can be used to find the equation of the perpendicular bisector?</h3>

The slope, <em>m</em>, of the line BC is calculated as follows;

  • m = (2 - 1)/(4 - (-2)) = 1/6

The slope of the perpendicular line to BC is -1/(1/6) = -6

The midpoint of the line BC is found as follows;

\left( - 2 +  \frac{4 - ( - 2)}{2}, \: 1 +  \frac{2 - 1}{2}   \right) = (1,\: 1.5)

The perpendicular bisector is the perpendicular line constructed from the midpoint of BC.

The equation of the perpendicular bisector in point and slope form is therefore;

(y - 1.5) = -6•(x - 1)

y - 1.6 = -6•x + 6

y = -6•x + 6 + 1.6 = 7.6 - 6•x

Which gives;

  • y = 7.6 - 6•x

Learn more about equations of perpendicular lines here:

brainly.com/question/11635157

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6 0
2 years ago
The line represents the number of packs of gum, y, remaining at a concession stand after x minutes.
Sav [38]

Answer:

B. the initial number of packs of gum

Step-by-step explanation:

Another name for y-intercept is initial value.

I am joyous to assist you anytime.

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3 years ago
Find a function ω(x) such that the function f(x) = sin(ω(x)) has the roots at π, π2
atroni [7]

So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.

f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}

8 0
3 years ago
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