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3 years ago

(2 x 1 1/6) + (1/2 x 1 1/6)

Mathematics

1 answer:

nadezda [96]3 years ago

5 0

Step-by-step explanation:

55/12 is your answer

please give me some thanks

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A big lockers above a smaller locker they both have 0.5 m wide and 0.6 m deep Big lockers 1.2 m tall and small lockers half of t

Tomtit [17]

Answer:

$0.54\,\,m^3$

Step-by-step explanation:

Given: Dimensions of a big locker are 0.5 m × 0.6 m × 1.2 m

Dimensions of a small locker are 0.5 m × 0.6 m × $\frac{1.2}{2}=0.6\,\,m$ (as height of small locker is half the height of big locker )

To find: total volume of one big locker and one small locker

Solution:

Volume of cuboid = length × breadth × height

Total volume of one big locker and one small locker = Total volume of one big locker + total volume of one small locker

= $0.5\times 0.6\times 1.2+0.5\times 0.6\times 0.6$

$=0.5\times 0.6\left ( 1.2+0.6 \right )\\=0.3(1.8)\\=0.54\,\,m^3$

4 0

3 years ago

Read 2 more answers

A triangle has base 15ft and area 60ft2. What is the height?j

hoa [83]

The height is 4 feet. You calculate the area by length times height. So what multiplied by 15 is 60? Well, 15ft x 4ft = 60ft^2

6 0

3 years ago

A kitten’s mass at birth was 0.09 kilogram. The kitten gained approximately 0.084 kilogram each week. After how many weeks is th

Naddika [18.5K]

A kitten’s mass at birth was 0.09 kilogram. The kitten gained approximately 0.084 kilogram each week. After how many weeks is the kitten’s mass 1.098 kilograms?

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2 years ago

Jane needs $20 to buy her radio. She has saved $15.

KonstantinChe [14]

Answer:

the answer is d

Step-by-step explanation:

$\frac{15}{20} \times 100 = 75$

7 0

3 years ago

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

5 0

3 years ago