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3 years ago

Cynthia clones 50 frogs in a year.

Mathematics

1 answer:

Dima020 [189]3 years ago

5 0

There are 52 weeks in a year. 1 frog per week: 52 * 1 = 52

So, cloning about 1 frog per week would be reasonable.

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Please answer me as soon as possible

adell [148]

Answer:????

Step-by-step explanation:

8 0

3 years ago

A jar made of 3/16-inch-thick glass has an inside radius of 3.00 inches and a total height of 6.00 inches (including the bottom

Nady [450]

Answer:

a) 0.012ft³

b)2.09 lb

c)0.791 lb

d) 0.012ft³

e) 6 inches

Step-by-step explanation:

a) To determine the volume of the shell of the jar we must determine the volume of the hollow cylinder:

$V=3.1416\cdot{h}\cdot{(R^2-r^2)$

Where r is the inside radius and R the outer radius.

$V=3.1416\cdot{6}\cdot{(3.1875^2-3^2)=21.87$

convert to cubic ft:

$=21.87\cdot{0.0005787}=0.01266$

The volume of the glass shell is 0.012 ft³

b) The weight of the cylinder can be determined because we have the volume and the density. Weight is defined as the product of volume and density:

$M=p\cdot{v}=165\cdot{0.01266}=2.09$

The weight of the jar is 2.09 lb

c) The displacement of water is just the density of water mutliplied by the volume of the jar:#

$=0.01266\cdot{62.5}=0.79$

0.791 lb of water will be displaced.

d) The volume of the water displaced is:

[tex]=0.791/62.5=0.01266

e) The jar will be fully submerged because the density of the jar is larger than water.

5 0

3 years ago

Round 36, 993 to nearest thousands

Natalija [7]

37,000 , The 6 is closer to 10 than it is 0

5 0

3 years ago

Read 2 more answers

Please help me solve this asap!

sveticcg [70]

The would be answer is A

3 0

3 years ago

NO LINKS OR FILES!

Archy [21]

(a) If the particle's position (measured with some unit) at time t is given by s(t), where

$s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}$

then the velocity at time t, v(t), is given by the derivative of s(t),

$v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}$

(b) The velocity after 3 seconds is

$v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}$

$\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}$

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

$\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|$

In interval notation, this happens for t in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt$

By definition of absolute value, we have

$|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)$

In part (d), we've shown that v(t) > 0 when -√11 < t < √11, so we split up the integral at t = √11 as

$\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt$

and by the fundamental theorem of calculus, since we know v(t) is the derivative of s(t), this reduces to

$s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}$

7 0

2 years ago