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2 years ago

15

The total weight of the fish tank in a tank of tropical fish at fish in fur was seven eights pounds. Each with 164 pounds. After

Eric bought some fish, the total Weight of the fish remaining in their tank was one and a half out how many fish did he buy? With exclamation

1 answer:

slavikrds [6]2 years ago

5 0

386 is the answer.................................

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Find the value of each variable

garri49 [273]

Triangles add to 180° and there is already 100° shown inside of the triangle (30° plus 70° = 100°). 180° -100° = 80°, which is x. X and y are vertical angles, so y is also 80°. The answer is B.

8 0

2 years ago

Is 74 fl oz greater than less than or equal to 8cups

Mariana [72]

It is greater than because 74 fl. oz. is equal to 9.25 cups.

6 0

2 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

2 years ago

kantabai bought 1/2 kg tea and 5 kg sugar from a shop .she paid rupess 50 as return fare for rickshaw .total expense was rupess

r-ruslan [8.4K]

Answer:

40 Rs per kg.

Step-by-step explanation:

solution

Let the rate of tea be x Rs per kg and that of sugar be y Rs per kg.

When Kantabai bought the items by going to the shop,

3/2x+5y+50=700

⇒3x+10y=1300....(I)

When Kantabai bought the items online then

2x+7y=880......(II)

Multiplying (I) with 2 and (II) with 3 we get

6x+20y=2600......(III)

6x+21y=2640.......(IV)

eq.(IV)−(III)

y=40

Putting the value of y =40 in (II)

2x+7(40)=880

⇒2x=880−280=600

⇒x=300

Thus, tea is at 300 Rs per kg and sugar is 40 Rs per kg.

8 0

1 year ago

A power plant has 4 tons of coal. a ton of coal produces 2,460 kilowatt hours of electricity. is this enough to power 9 light bu

igomit [66]

Assuming 100 kw bulb

<span>We'll start by figuring out how much energy in kilowatt-hours the light bulb uses per year. We multiply how much power it uses in kilowatts, by the number of hours in a year. That gives 0.1 kW x 8,760 hours or </span>876 kWh x 9 bulbs = 7884 kWh

The electricity produced by 4 tons x 2460 kWh/ton = 9840 kwh

<span>Since the produced electricity is bigger so it is enough</span>

5 0

3 years ago