All categories

2 years ago

Estimate the sum of 1,256, 379 and 305,986 by first rounding each number to the nearest ten thousand.

Mathematics

1 answer:

KatRina [158]2 years ago

3 0

9514 1404 393

Answer:

1,570,000

Step-by-step explanation:

The sum is approximately ...

1,260,000 +310,000 = 1,570,000

_____

<em>Additional comment</em>

It is a good idea to estimate the error associated with an estimate. Here, both numbers are rounded up by about 4000 each, so the estimate is around 8000 high.

You might be interested in

Find the area of the trapezoid

Yuliya22 [10]

The area of a trapezoid is half its height multiplied by the sum of its two bases. So in other words 1/2h(b1+b2)

5 0

3 years ago

Read 2 more answers

If a school policy maker wanted to learn what expectations must be met to receive funding from H.R.5, what titles would that per

lukranit [14]

I'm definite about a b c

7 0

3 years ago

0.4 times 0.59<br> sgaehags

ra1l [238]

Answer:

0.236

Step-by-step explanation:

Multiply 0.4 by 0.59

hopes this helps have a good day :) <3

brainlest???

5 0

2 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

i promise i will mark brainliest if the point (-3,4) is the vertex point of the curve of aquadratic function f and -5 is aroot o

Jet001 [13]

Ans:

(x+5)(x+1)+8

Step by step explanation:

-5 was mentioned as the 0 of the first root: (x+5). Then with some maneuvering we can plot (-3,4) then add (x+1) because they land on the same y axis. then we add 8 to get it right on the vertex.

6 0

3 years ago