1.8 + 45 tenths = __

For the following sequence, determine if it is an arithmeticsequence, a geometric sequence, or neither. If it is either arith

uysha [10]

Answer: An arithmetic sequence is of the form:

aₙ = a₀ + (n -1)*d

So the difference between a term and the nextone is always the same, and in our case the numbers are 5,8,13,21,34,55.

The distance between the numbers is not constant, so this is not a arithmetic sequence.

a geometric sequence is of the form: aₙ = a*rⁿ

So the therms grow exponentially.

here a₀ = a*r⁰ = a = 5.

a₁ = a*r¹ = 5*r = 13, then r = 13/5 = 2.6

a₂ = a*r² = 5*2.6*2.6 = 33.8, so here we can see that this is not our series, so our series isn't geometric.

So the answer for the first part is neither.

Now, the sequence is 5,8,13,21,34,55, ...

you can see that

5 + 8 = 13

13 + 8 = 21,

21 + 34 = 55

So our sequence is of the form: aₙ = aₙ₋₁ + aₙ₋₂.

then, the next term will be: 55 + 21= 76.

4 0

3 years ago

Approximate area under the curve f(x) =-x^2+2x+4 from x=0 to x=3 by using summation notation with six rectangles and use the the

bekas [8.4K]

Answer:

Summation notation:

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

or after using your function part:

$\frac{1}{2}\sum_{k=1}^6(-(.5k)^2+2(.5k)+4)$

After evaluating you get 11.125 square units.

Step-by-step explanation:

The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.

The distance between x=0 and x=3 is (3-0)=3.

We want to divide that length of 3 units by 6 which gives a length of a half per each base length.

We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).

The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).

The next will be at x=2.5-.5=2, and the third rectangle will have a height of f(2).

The fourth rectangle will have a height of f(2-.5)=f(1.5).

The fifth one will have a height of f(1.5-.5)=f(1).

The last one because it is the sixth one will have a height of f(1-.5)=f(.5).

So to find the area of a rectangle you do base*time.

So we just need to evaluate:

$\frac{1}{2}f(3)+\frac{1}{2}f(2.5)+\frac{1}{2}f(2)+\frac{1}{2}f(1.5)+\frac{1}{2}f(1)+\frac{1}{2}f(.5)$

or by factoring out the 1/2 part:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

To find f(3) replace x in -x^2+2x+4 with 3:

-3^2+2(3)+4

-9+6+4

1

To find f(2.5) replace x in -x^2+2x+4 with 2.5:

-2.5^2+2(2.5)+4

-6.25+5+4

2.75

To find f(2) replace x in -x^2+2x+4 with 2:

-2^2+2(2)+4

-4+4+4

4

To find (1.5) replace x in -x^2+2x+4 with 1.5:

-1.5^2+2(1.5)+4

-2.25+3+4

4.75

To find f(1) replace x in -x^2+2x+4 with 1:

-1^2+2(1)+4

-1+2+4

5

To find f(.5) replace x in -x^2+2x+4 with .5:

-.5^2+2(.5)+4

-.25+1+4

4.75

Now let's add those heights. After we obtain this sum we multiply by 1/2 and we have our approximate area:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

$\frac{1}{2}(1+2.75+4+4.75+5+4.75)$

$\frac{1}{2}(22.25)$

$11.125$

Okay now if you wanted the summation notation for:

$\frac{1}{2}(f(3)+f(2.5)+f(2)+f(1.5)+f(1)+f(.5))$

is it

$\frac{1}{2}\sum_{k=1}^{6}(f(.5+.5(k-1)))$

or after simplifying a bit:

$\frac{1}{2}\sum_{k=1}^6 f((.5+.5k-.5))$

$\frac{1}{2}\sum_{k=1}^6f((.5k))$

If you are wondering how I obtain the .5+.5(k-1):

I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)

6 0

3 years ago

Solve the equation using square roots. (X+3)^2=0

Vilka [71]

Answer:

x=3

Step-by-step explanation:

(x−3)^2=0

Set the x−3 equal to 0.

x−3=0

Add 3 to both sides of the equation.

x=3

7 0

2 years ago

Other expressions for 19 times 25 + 19 times 75

Vladimir [108]

Answer:

1,900

Step-by-step explanation:

4 0

3 years ago

If 5 notebooks cost $5.25 how much is 3 notebooks

mrs_skeptik [129]

3 notebooks will cost $3.15

6 0

2 years ago

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1.8 + 45 tenths = ___