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Mathematics

2 answers:

denpristay [2]3 years ago

7 0

7,4 ×10^ -5

.....................

iVinArrow [24]3 years ago

6 0

The second option, fourth, and the last are all scientific notation.

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The weight of an object on the moon varies directly with its weight on the earth. If an object weighing 95 lbs on the moon weigh

Luda [366]

ANSWER

450lb

EXPLANATION

If the weight, m of an object on the moon varies directly as the weight of an object e, on the earth, then we can write the mathematical statement,.
$m \propto \: e$
We introduce the constant of proportionality to obtain,

$m = ke$

When, m=95, e=570,

This implies that,

$95 = 570k$
$k = \frac{95}{570}$

$k = \frac{1}{6}$

The equation now becomes,

$m = \frac{1}{6} e$

We want to find e, when m=2700,

$m = \frac{1}{6} \times 2700$

$m =450lb$

8 0

3 years ago

Following decimals into words 15.06

melamori03 [73]

Answer:

Fifteen and six hundredths

Step-by-step explanation:

6 0

3 years ago

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Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$