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3 years ago

Rajesh is working out the area of a circle with radius 10 he writes

Mathematics

1 answer:

Maru [420]3 years ago

7 0

Answer:

He multiplied the radius by 2 instead of squaring it

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Help please i need answers fast if you help or try to help tysm *hearts*

Tanzania [10]

Answer:

C. 3rd one Fraction 10/24

Step-by-step explanation:

3/4 - 2/6 = 0.41 or = 10/24 or it could be equal to 5/12 are 10/24, 15/36, 20/48 and 25/60 or as Decimal 0.41

Step-by-step explanation:

Hope This Helped

7 0

1 year ago

an ostrich egg weighs 2.9 pounds.The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds.write

Alex73 [517]

Answer:

$2.9-w=1.6$

Step-by-step explanation:

Let w be weight of emu egg in pounds.

We have been given that an ostrich egg weighs 2.9 pounds.The difference between the weight of this egg and the weight of an emu egg is 1.6 pounds.

We can represent this information as:

$2.9-w=1.6$

Therefore, the equation $2.9-w=1.6$ will give us the weight of the emu egg, w, in pounds.

8 0

3 years ago

Read 2 more answers

HELP PLEASE I NEED ANSWER NOW

Rainbow [258]

Answer:B

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

The Volume of a sphare is 28/3 times the surface area calculate The surface area and the Volume of the sphere, correct to the ne

Aloiza [94]

Given:

Volume of a sphere is $\dfrac{28}{3}$ times the surface area.

To find:

The surface area and the volume of the sphere.

Solution:

Volume of a sphere:

$V=\dfrac{4}{3}\pi r^3$ ...(i)

Surface area of a sphere:

$A=4\pi r^2$ ...(ii)

Where, r is the radius of the sphere.

Volume of a sphere is $\dfrac{28}{3}$ times the surface area.

$V=\dfrac{28}{3}\times A$

$\dfrac{4}{3}\pi r^3=\dfrac{28}{3}\times 4\pi r^2$

Multiply both sides by 3.

$4\pi r^3=112\pi r^2$

$\dfrac{\pi r^3}{\pi r^2}=\dfrac{112}{4}$

$r=28$

Using (i), the volume of the sphere is:

$V=\dfrac{4}{3}\times \dfrac{22}{7}\times (28)^3$

$V\approx 91989$

Using (ii), the surface area of the sphere is:

$A=4\times \dfrac{22}{7}\times (28)^2$

$A=9856$

Therefore, the surface area of the sphere is 9856 sq. units and the volume of the sphere is 91989 cubic units.

3 0

3 years ago

A rectangle initially has dimensions 6 cm by 7 cm. All sides begin increasing in length at a rate of 1 cm/s. At what rate is the

Morgarella [4.7K]

Answer:

49 cm²/s

Step-by-step explanation:

The problem statement tells you what to do.

Write an equation relating A, b, h:

A = bh . . . . . . the equation for the area of a rectangle

Differentiate with respect to t:

dA/dt = (db/dt)h + b(dh/dt) . . . . . . . product rule

To find the rate of change after 18 seconds, you need to know the dimensions b and h after 18 seconds. Since each dimension was increasing at the rate of 1 cm/s, it is 18 cm more than it was at the beginning:

At 18 seconds,

b = 6 cm + 18 cm = 24 cm;

h = 7 cm + 18 cm = 25 cm.

Of course, db/dt = dh/dt = 1 cm/s. Then the rate of change of area is ...

dA/dt = (1 cm/s)(25 cm) + (24 cm)(1 cm/s)

dA/dt = 49 cm²/s

_____

You could write a formula for the area as a function of time and differentiate that:

A = (6 +t)(7 +t) = 42 + 13t + t²

Then the derivative is ...

dA/dt = 13 +2t

and when t=18, this is ...

dA/dt = 13 + 2(18) = 49 . . . . cm²/s

7 0

3 years ago