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2 years ago

12

Estimate the sum of the decimals below by rounding to the nearest whole

1 answer:

Tamiku [17]2 years ago

7 0

Answer:

13

Step-by-step explanation:

to round to nearest whole number if its .5 or above you go up 1. If it's .4 or below you go down 1.

2.414=2

8.160=8

2.621=3

add

2+8+3=13

if you just added the decimals you would get: 13.195. (This shows you have a good estimate.)

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What is the measure of А<br> 50"<br> 50°<br> B<br> C<br> [Not drawn to scale]<br> 50°<br> 80<br> 100<br> 130

Paladinen [302]

Answer:

solution given :

<A=?

<B=50°

<C=50°

if it is a triangle:

<A+<B+<C=180°( sum of interior angle of a triangle is 180°)

<A=180-50-50

<A=80°

<<u>A=</u><u>8</u><u>0° is a required answer.</u>

7 0

3 years ago

Plllzzz im new and i neeed help

ella [17]

Answer:

4082

Step-by-step explanation:

Given

The composite object

Required

The volume

The object is a mix of a cone and a hemisphere

Such that:

<u>Cone</u>

$r = 10cm$ ---- radius (r = 20/2)

$h = 19cm$

<u>Hemisphere</u>

$r=10cm$

The volume of the cone is:

$V_1 = \frac{1}{3}\pi r^2h$

$V_1 = \frac{1}{3}\pi * 10^2 * 19$

$V_1 = \frac{1900}{3}\pi$

The volume of the hemisphere is:

$V_2 = \frac{2}{3}\pi r^3$

$V_2 = \frac{2}{3}\pi 10^3$

$V_2 = \frac{2000}{3}\pi$

So, the volume of the object is:

$V = V_1 + V_2$

$V = \frac{1900}{3}\pi + \frac{2000}{3}\pi$

$V = \frac{3900}{3}\pi$

$V = 1300\pi$

$V = 1300 * 3.14$

$V = 4082$

8 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

So

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

3 years ago

Plzz help will mark brainliest don't do it for the points plzz!!!!

masya89 [10]

Answer: y=2x

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

17. There are 25 squares in the pattern.

GrogVix [38]

A) 14/25
b) 4/25
c) 2/25

6 0

3 years ago