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anygoal [31]
3 years ago
15

Find the Perimeter Of these shapes using Pythagorean theory

Mathematics
1 answer:
Ludmilka [50]3 years ago
4 0
1 is 47 and 2 is 27 you welcome
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Please help! Will mark the brainliest!
dimaraw [331]
(f+g)(5)=33 is the answer
3 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
The area of a rhombus is 90 square inches. If the longer diagonal measures 18 inches, what is the length of the shorter diagonal
lorasvet [3.4K]
10 Inches

A = p q / 2

p = diagonal 1
q = diagonal 2

2A / q = p

Solve for p
180/18 = 10
5 0
3 years ago
Angad, Caroline and Sarah share some sweets in the ratio 7:4:3. Angad gets 52 more sweets than Sarah. How many sweets does Carol
padilas [110]
A = 7x, c = 4x, s = 3x

a = s+52
7x = 3x+52
4x = 52
x = 13

c = 4x = 52
Caroline gets 52.
7 0
3 years ago
24 over 32 as an equivelent fraction
andrew-mc [135]
12/16,
6/8,
3/4, and
45/60

are all equivalent to 24/32.

Now that I've listed those four, there are
only an infinite number of others remaining.
4 0
3 years ago
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