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3 years ago

What is the sum of 3.6 × 103 and 6.1 × 103?

Mathematics

2 answers:

ahrayia [7]3 years ago

5 0

999.1 is the answer you multiply both then add

LiRa [457]3 years ago

3 0

The answer is 999.1. because of Order of Operations, you can rewrite this problem as (3.6+6.1)*103. you can add what is in the parenthesis then multiply the answer by 103.

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A trough is 5 meters long, 2 meters wide, and 3 meters deep. The vertical cross-section of the trough parallel to an end is shap

SSSSS [86.1K]

We need a work of 294210 watts to pump the water over the top. $\blacksquare$

<h3>Work needed to pump all the water over the top</h3>

Since the <em>cross section</em> area of the trough ( $A$ ), in square meters, varies with the height of the water ( $h$ ), in meters, and considering that <em>pumping</em> system extracts water at <em>constant</em> rate, then the work needed to pump all the water ( $W$ ), in joules, is:

$W = \int\limits^{V_{max}}_0 {p} \, dV$ (1)

Where:

$p$ - Pressure of the infinitesimal volume, in pascals.
$V$ - Volume, in cubic meters.
$V_{max}$ - Maximum volume allowed by the trough, in cubic meters.

The <em>infinitesimal</em> volume is equivalent to the following expression:

$dV = A\, dh$ (2)

Since the area is directly proportional to the height of the water, we have the following expression:

$A = \frac{A_{max}}{H_{max}}\cdot h$ (3)

Where:

$A_{max}$ - Area of the base of the trough, in square meters.
$H_{max}$ - Maximum height of the water, in meters.

In addition, we know that pressure of the water is entirely hydrostatic:

$p = \rho \cdot g \cdot h$ (4)

Where:

$\rho$ - Density of water, in kilograms per cubic meters.
$g$ - Gravitational acceleration, in meters per square second.

By (2), (3) and (4) in (1):

$W = \frac{\rho\cdot g\cdot W_{max}\cdot L_{max}}{H_{max}} \int\limits^{H_{max}}_{0} {h^{2}} \, dh$ (5)

Where:

$W_{max}$ - Width of the base of the triangle, in meters.
$L_{max}$ - Length of the base of the triangle, in meters.
$H_{max}$ - Maximum height of the triangle, in meters.

The resulting expression is:

$W = \frac{\rho\cdot g\cdot W_{max}\cdot L_{max}\cdot H_{max}^{2}}{3}$ (5b)

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $W_{max} = 2\,m$ , $L_{max} = 5\,m$ and $H_{max} = 3\,m$ , then the work needed to pump the water is:

$W = \frac{(1000)\cdot (9.807)\cdot (2)\cdot (5)\cdot (3)^{2}}{3}$

$W = 294210\,W$

We need a work of 294210 watts to pump the water over the top. $\blacksquare$

To learn more on work, we kindly invite to check this verified question: brainly.com/question/17290830

6 0

2 years ago

Eva and Liam are reading a novel in their English class Liam has read 17 less pages than Eva has if they read 265 pages combined

lara [203]

Eva has read 141 pages and Liam has read 124 pages.

Step-by-step explanation:

Let,

Pages read by Eva = x

Pages read by Liam = y

According to given statement;

x+y=265 Eqn 1

y=x-17 Eqn 2

Putting value of y from Eqn 2 in Eqn 1

$x+(x-17)=265\\x+x-17=265\\2x-17=265\\2x=265+17\\2x=282$

Dividing both sides by 2

$\frac{2x}{2}=\frac{282}{2}\\x=141$

Putting x=141 in Eqn 2

$y=141-17\\y=124$

Eva has read 141 pages and Liam has read 124 pages.

Keywords: Linear equation, substitution method

Learn more about linear equations at:

#LearnwithBrainly

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3 years ago

If the point (−1, −5) is reflected across the x-axis, what is the location of the new point? A) (5, 1) B) (1, 5) C) (1, −5) D) (

olchik [2.2K]

Answer:

D (-1, 5)

Step-by-step explanation:

In reflection along the x-axis,

x-coordonate remains unchanged and y-coordinate gets multiplied by -1

4 0

3 years ago

One number is four more than twice another. Their sum<br> is 34. Find the numbers.

lianna [129]

x = 15

2x + 4 = 34

-4 -4

2 2

X = 15

5 0

3 years ago

In Exploration 5.4.2 Question 2, what conclusion can you make about the value of the derivative at

givi [52]

The value of the derivative at the maximum or minimum for a continuous function must be zero.

<h3>What happens with the derivative at the maximum of minimum?</h3>

So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.

Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).

If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.

So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.

If you want to learn more about maximums and minimums, you can read:

brainly.com/question/24701109

4 0

2 years ago