This sequence be represented as a recursive equation by a1=8 and an=2a1
<u>Step-by-step explanation</u>:
- 'Recursive' refers to the repetition of a specific process in a sequence.
- The given sequence is {8,16,32,64}.
- If the value is 2 times the previous value, then an=2a(n-1)
Let a1=8,
then a2 = 2a(2-1)
⇒ a2 = 2a1
⇒ a2 = 2(8)
⇒ a2 = 16
Similarly,
For a2=16,
⇒ a3 = 2(a2)
⇒ a3 = 2(16)
⇒ a3 = 32
For a3=32,
⇒ a4 = 2(a3)
⇒ a4 = 2(32)
⇒ a4 = 64
∴ The equation is recursive as a1=8 and an=2a1 to follow the sequence.
The answer is A !!!!!!!!
the two segments AD and AB are equal therefore if you make the two equal to each other you will get the answer:
<span>x^2+2=11 then you subtract two to put x squared on the side by itself
</span>then you have <span>x^2=9
then all you have to do is take the square root of 9 and </span><span>x^2 and you get x=3
thefore the answer is A</span>
Answer:
perimeter is 4 sqrt(29) + 4pi cm
area is 40 + 8pi cm^2
Step-by-step explanation:
We have a semicircle and a triangle
First the semicircle with diameter 8
A = 1/2 pi r^2 for a semicircle
r = d/2 = 8/2 =4
A = 1/2 pi ( 4)^2
=1/2 pi *16
= 8pi
Now the triangle with base 8 and height 10
A = 1/2 bh
=1/2 8*10
= 40
Add the areas together
A = 40 + 8pi cm^2
Now the perimeter
We have 1/2 of the circumference
1/2 C =1/2 pi *d
= 1/2 pi 8
= 4pi
Now we need to find the length of the hypotenuse of the right triangles
using the pythagorean theorem
a^2+b^2 = c^2
The base is 4 ( 1/2 of the diameter) and the height is 10
4^2 + 10 ^2 = c^2
16 + 100 = c^2
116 = c^2
sqrt(116) = c
2 sqrt(29) = c
Each hypotenuse is the same so we have
hypotenuse + hypotenuse + 1/2 circumference
2 sqrt(29) + 2 sqrt(29) + 4 pi
4 sqrt(29) + 4pi cm
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Define x :
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Let the smallest number be x.
The other two numbers will be x+1 and x+2.
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Construct equation :
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The sum of all 3 numbers is 78
⇒ x + x + 1 + x + 2 = 78
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Solve x :
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x + x + 1 + x + 2 = 78
3x + 3 = 78
3x = 78 - 3
3x = 75
x = 25
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Answer: Smallest number = 25
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