Answer:
4.5 sq. units.
Step-by-step explanation:
The given curve is 
⇒ 
...... (1)
This curve passes through (0,0) point.
Now, the straight line is y = 3x - 6 ....... (2)
Now, solving (1) and (2) we get,
⇒ (y - 3)(y + 2) = 0
⇒ y = 3 or y = -2
We will consider y = 3.
Now, y = 3x - 6 has zero at x = 2.
Therefor, the required are = 
= ![\sqrt{3} [{\frac{x^{\frac{3}{2} } }{\frac{3}{2} } }]^{3} _{0} - [\frac{3x^{2} }{2} - 6x ]^{3} _{2}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%5B%7B%5Cfrac%7Bx%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%7D%5D%5E%7B3%7D%20_%7B0%7D%20-%20%5B%5Cfrac%7B3x%5E%7B2%7D%20%7D%7B2%7D%20-%206x%20%5D%5E%7B3%7D%20_%7B2%7D)
= ![[\frac{\sqrt{3}\times 2 \times 3^{\frac{3}{2} } }{3}] - [13.5 - 18 - 6 + 12]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Csqrt%7B3%7D%5Ctimes%202%20%5Ctimes%203%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B3%7D%5D%20-%20%5B13.5%20-%2018%20-%206%20%2B%2012%5D)
= 6 - 1.5
= 4.5 sq. units. (Answer)
The area of the resulting figure will be given by:
∫f(x)dx
f(x)=13/2x^3
thus
∫f(x)dx=13/2∫x³dx=13/8[x^4]
integrating over the inerval
13/8(12^4)-13/8(5^4)
=32680+3/8 sq. units
=
First replace y with 15.
1/5x-10=30
Then just solve from there.
x=200
Answer:
Step-by-step explanation:
The distance formula states that the distance between two points 
and 
is 
.
The two points we have are 
and 
. Plugging these numbers into the distance formula, we have
.
Simplifying with order of operations, first using the distributive property, gives
.
Squaring and adding gives
which is the answer in simplest form. This also rounds to about 12.04.
A'(-6, -10), B'(-3,-13), and C'(-5,-1) are the vertices of the ΔA'B'C' under the translation rule (x,y)→(x,y-3). This can be obtained by putting the ΔABC's vertices' values in (x, y-3).
<h3>Calculate the vertices of ΔA'B'C':</h3>
Given that,
ΔABC : A(-6,-7), B(-3,-10), C(-5,2)
(x,y)→(x,y-3)
The vertices are:
- A(-6,-7 )⇒ (-6,-7-3) = A'(-6, -10)
- B(-3,-10) ⇒ (-3,-10-3) = B'(-3,-13)
- C(-5,2) ⇒ (-5,2-3) = C'(-5,-1)
Hence A'(-6, -10), B'(-3,-13), and C'(-5,-1) are the vertices of the ΔA'B'C' under the translation rule (x,y)→(x,y-3).
Learn more about translation rule:
brainly.com/question/15161224
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