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Tasya [4]
4 years ago
11

How should some one classify the water 2.025 gram 457 mg 589 mg

Mathematics
1 answer:
Juliette [100K]4 years ago
7 0
Grams to what? 
milligrams to what? 

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A child's sandbox has a length of 5 metres and a width of 4 metres. A uniform sidewalk is to be built around the sandbox with pa
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Given that 1 x2 dx 0 = 1 3 , use this fact and the properties of integrals to evaluate 1 (4 − 6x2) dx. 0
Debora [2.8K]

So, the definite integral  \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Given that

\int\limits^1_0 {x^{2} } \, dx = 13

We find

\int\limits^1_0 {(4 - 6x^{2} )} \, dx

<h3>Definite integrals </h3>

Definite integrals are integral values that are obtained by integrating a function between two values.

So, Integral \int\limits^1_0 {(4 - 6x^{2} )} \, dx

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx = \int\limits^1_0 {4} \, dx - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - \int\limits^1_0 {6x^{2} } \, dx \\=  4[x]^{1}_{0}    - 6\int\limits^1_0 {x^{2} } \, dx \\= 4[1 - 0]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4[1]    - 6\int\limits^1_0 {x^{2} } \, dx\\= 4    - 6\int\limits^1_0 {x^{2} } \, dx

Since

\int\limits^1_0 {x^{2} } \, dx = 13,

Substituting this into the equation the equation, we have

\int\limits^1_0 {(4 - 6x^{2} )} \, dx = 4 - 6\int\limits^1_0 {x^{2} } \, dx\\= 4 - 6 X 13 \\= 4 - 78\\= -74

So, \int\limits^1_0 {(4 - 6x^{2} )} \, dx= - 74

Learn more about definite integrals here:

brainly.com/question/17074932

4 0
3 years ago
In the diagram below, line AC intersects triangle BDE at B. Choose True or False for each statement.
mixer [17]

Answer:

t'

f

t

Step-by-step explanation:

6 0
3 years ago
Find an equation of the sphere that passes through the origin and whose center is (-2, 2, 3). Be sure that your formula is monic
Andrei [34K]

Answer:

\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}

Step-by-step explanation:

Given the center of sphere is: (-2, 2, 3)

Passes through the origin i.e. (0, 0, 0)

To find:

The equation of the sphere ?

Solution:

First of all, let us have a look at the equation of a sphere:

(x-a)^2+(y-b)^2+(z-c)^2=r^2

Where (x,y,z) are the points on sphere.

(a, b, c) is the center of the sphere and

r is the radius of the sphere.

Radius of the sphere is nothing but the distance between any point on the sphere and the center.

We are given both the points, so we can use distance formula to find the radius of the given sphere:

D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}

Here,

x_1 =0 \\y_1 =0 \\z_1 =0 \\x_2 =-2 \\y_2  =2 \\z_2 =3

So, Radius is:

r = \sqrt{(-2-0)^2+(2-0)^2+(3-0)^2}\\\Rightarrow r = \sqrt{4+4+9} = \sqrt{17}

Therefore the equation of the sphere is:

(x-(-2))^2+(y-2)^2+(z-2)^2=(\sqrt{17})^2\\\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}

4 0
4 years ago
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