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2 years ago

A room 9.5 m long and 6 m wide is surrounded by a verandah 1.25 m wide. Calculate the cost of

Mathematics

1 answer:

oksano4ka [1.4K]2 years ago

5 0

Answer:

Step-by-step explanation:

Area of room = 6×9.5=57m2

Area of room including verandah = 12×8.5=102m2

So area of verandah = 102−57=45m2

So cost of cementing the floor of this verandah = 45×80=3600rupees

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Help asap!!!!!!Order the number below from least to greatest

OLEGan [10]

Answer:

Step-by-step explanation:

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2 years ago

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Mr. Schmidt is making strawberry jam. He fills 31-pint jars and 13-quart jars. Then he gives 20 quarts of jam to his friends. Ho

MArishka [77]

1 quart= 2 pints so 31 pints = 15.5 quarts so 15.5+13=23.5-20= 3.5 left hope this explains it better!!

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2 years ago

What are the laws of exponents for class 7

finlep [7]

Answer:

Multiplication of no. with same base

division of no. with same base

taking power of a power

multiplication of two numbers with same exponents

division of two numbers with same exponents

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Step-by-step explanation:

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3 years ago

9x2+4y2 = 36

valentinak56 [21]

Answer:

The foci of the given ellipse = (0 ,-√5 ) and ( 0, √5 )

Step-by-step explanation:

Step(i):-

Given Ellipse 9 x² + 4 y² = 36

$\frac{9x^{2} }{36} + \frac{4y^{2} }{36} =1$

⇒ $\frac{x^{2} }{4} + \frac{y^{2} }{9} = 1$

Step(ii):-

The Major axis lies on y-axis

Foci of the Formula C² = a² -b² = 9-4 =5

C = √5

The Foci is always lie on the major axis(longest) axis, spaced equally each side of the centre

Given ellipse of the foci is lie on Y- axis

Final answer:-

The foci of the given ellipse = (0 ,-√5 ) and ( 0, √5 )

8 0

2 years ago

A tank initially contains 50 liters of water with 15kg of salt mixed in. Salt water with concentration 1/3 kg/L runs in at the r

Nimfa-mama [501]

Answer:

13.57 kg

Step-by-step explanation:

Let the amount of salt in a tank be = $\frac{salt in the tank}{volume of salt in the tank}$

= $\frac{15}{50}$

= 0.3 kg/L

The concentration of salt coming into the tank = 0.333kg/L

We know that the concentration of salt leaving the tank should be:

amount of salt in the tank/volume of tank = y(t)/50 L

rate of change of salt = rate in - rate out

$\frac{dy}{dt}$ = 0.66 - $\frac{y}{50}$

solving the differential equation gives :

ln y = 0.66/y (t) - 0.02 t + C

Initial conditions, t= 0; y (0) = 15

then ln 15 = C

ln y = $\frac{0.66}{y}$ t - 0.02t + ln 15

ln $\frac{y}{15}$ = 0.02t

y = 15e( exp -0.02t)

the amount of salt in 5 mins will be:

y (t) = 15e(exp -0.02 (5)

= 13.57 kg

4 0

2 years ago