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3 years ago

HELP ME

Mathematics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

A.20 ft

Step-by-step explanation:

the shadow is double the height of the object.

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Find the centroid of the region bounded by y = 3x^2, y = 0, x = 4

ANTONII [103]

To determine the centroid, we use the equations:

x⁻ = 1/A (∫ (x dA))
y⁻ = 1/A (∫ (y dA))

First, we evaluate the value of A and dA as follows:

A = ∫dA
A = ∫ydx
A = ∫3x^2 dx
A = 3x^3 / 3 from 0 to 4
A = x^3 from 0 to 4
A = 64

We use the equations for the centroid,
x⁻ = 1/A (∫ (x dA))
x⁻ = 1/64 (∫ (x (3x^2 dx)))
x⁻ = 1/64 (∫ (3x^3 dx)
x⁻ = 1/64 (3 x^4 / 4) from 0 to 4
x⁻ = 1/64 (192) = 3

y⁻ = 1/A (∫ (y dA))
y⁻ = 1/64 (∫ (3x^2 (3x^2 dx)))
y⁻ = 1/64 (∫ (9x^4 dx)
y⁻ = 1/64 (9x^5 / 5) from 0 to 4
y⁻ = 1/64 (9216/5) = 144/5

The centroid of the curve is found at (3, 144/5).

7 0

3 years ago

Reduce 6/12 to its simplest form

Vlad1618 [11]

Answer:

1/2

Step-by-step explanation:

6/12

3/6

1/2

7 0

3 years ago

Read 2 more answers

A shipping crate holds 14 books. The dimensions of each book are 4 inches by 11 by 14 inches. For A-C, select Yes or No to indic

Inga [223]

Answer:

A. No

B. Yes

C. No

Step-by-step explanation:

Dimensions of each book are 4 inches by 11 by 14 inches.

A. Each book has a volume of 29 cubic inches.

Volume of each book = length × width × height

= 4 × 11 × 14

= 616 cubic inches

B. If the crate could hold 18 books the volume of the crate would be 11,088.

Volume of each book = 616 cubic inches

Volume of the crate = volume of each book × number of books

= 616 × 18

= 11,088 cubic inches

Yes

C. Each crate has a volume of about 8,524 cubic inches

If the crate holds 14 books,

Volume of the crate = volume of each book × number of books

= 616 × 14

= 8,624 cubic inches

5 0

3 years ago

PLEASE HELP ILL MARK BRAINLIEST !!!

Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

5 0

3 years ago

78+24-11+56+23-117+1

kakasveta [241]

78+24+11+56+23+117+1 =310

4 0

3 years ago