Answer:
The sequence of first five terms
4 , 5, 6, 7, 8
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given
of the sequence
aₙ = n + 3 ..(i)
put n =1
a₁ = 1 +3
a₁ = 4
The first term of the sequence = 4
Put n =2 in equation (i)
a₂ = 2+3
a₂ = 5
The second term of the sequence = 5
put n=3 in equation (i)
a₃ = 3+3 = 6
The third term of the sequence = 6
Put n=4 in equation (i)
a₄ = 4 +3
a₄ = 7
The fourth term of the sequence = 7
Put n = 5 in equation (i)
a₅ = 5+3
a₅ = 8
The fifth term of the sequence = 8
<u><em>Step(ii):-</em></u>
The sequence of first five terms
4 , 5, 6, 7, 8
Answer:
132::
Step-by-step explanation:
The power of g00gle <3
Answer:
-1.875
Step-by-step explanation:
1. Write the problem: -2(5x + 8) = 14 + 6x
2. Multiply -2 by 5 and 8 to get: -10x - 16 = 14 + 6x
3. Combine like terms: -16x - 16 = 14
4. Repeat step 3: -16x = 30
5. Divide -16 by 30 to get the answer which is -1.875.
Hope this helped!
Part A would be the second option, as well as part B
Let
x--------> the length side of the square base
h--------> the height of the box
we know that
<u>the volume of the box is equal to</u>
![V=x^{2} *h\\ V=9\ m^{3}](https://tex.z-dn.net/?f=V%3Dx%5E%7B2%7D%20%2Ah%5C%5C%20V%3D9%5C%20m%5E%7B3%7D)
so
![x^{2} *h=9\\\\h=\frac{9}{x^{2} }](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2Ah%3D9%5C%5C%5C%5Ch%3D%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%20%7D)
<u>the surface area of the box is equal to</u>
(remember that the box is open)
area of the base=![(x^{2})\ m^{2}](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%29%5C%20m%5E%7B2%7D)
Perimeter of the base=![(4*x)\ m](https://tex.z-dn.net/?f=%284%2Ax%29%5C%20m)
height=(h) m
![h=\frac{9}{x^{2}}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%7D)
substitute
![SA=x^{2} +4*x*h\\ \\ \\ SA=x^{2} +4*x*\frac{9}{x^{2} } \\ \\ SA=x^{2} +\frac{36}{x}](https://tex.z-dn.net/?f=SA%3Dx%5E%7B2%7D%20%2B4%2Ax%2Ah%5C%5C%20%5C%5C%20%5C%5C%20SA%3Dx%5E%7B2%7D%20%2B4%2Ax%2A%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%20%7D%20%5C%5C%20%5C%5C%20SA%3Dx%5E%7B2%7D%20%2B%5Cfrac%7B36%7D%7Bx%7D)
we know that
the value of x can not be negative and the denominator can not be zero
therefore
<u>the answer is</u>
the domain of SA is x> 0
the domain is the interval-------------> (0,∞)