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alukav5142 [94]
3 years ago
11

What is the perimeter of a trapezoid that is 9.85 15 11 9

Mathematics
2 answers:
Alenkasestr [34]3 years ago
5 0
So you would add the numbers together:

9.85+15+11+9=44.85

So the perimeter is 44.85
DedPeter [7]3 years ago
3 0

Answer: 44.85 units

Step-by-step explanation: In this problem, we are asked to find the perimeter of the trapezoid. Remember that perimeter is the distance around a figure. We can find the perimeter of the trapezoid by adding the lengths of each of the sides.

9.85 + 15 + 11 + 9 = 44.85 units  

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What is the area of figure abcd? a trapezoid abcd is drawn with length of parallel sides ab and cd equal to 10 inches and 12 inc
Mama L [17]

Answer: 60 square inches

Step-by-step explanation:

4 0
3 years ago
Jerome, kevin and seth shared a submarine sandwhich. Jerome ate 1/2 of the sandwich , kevin ate 1/3 of the sandwich, and seth at
Paul [167]

Given:

Jerome ate 1/2 of the sandwich , Kevin ate 1/3 of the sandwich, and Seth ate the rest.

To find:

The ratio of jerome's share to kevin's share to seth's share.

Solution:

Jerome ate 1/2 of the sandwich. So,

Jerome's share = \dfrac{1}{2}

Kevin ate 1/3 of the sandwich. So,

Kevin's share = \dfrac{1}{3}

Seth ate the rest. So,

Seth's share = 1-\dfrac{1}{2}-\dfrac{1}{3}

                     = \dfrac{6-3-2}{6}

                    = \dfrac{1}{6}

The ratio of jerome's share to kevin's share to seth's share is:

\dfrac{1}{2}:\dfrac{1}{3}:\dfrac{1}{6}

To simplify the ratio multiply the ratio by 6.

\dfrac{1}{2}\times 6:\dfrac{1}{3}\times 6:\dfrac{1}{6}\times 6

3:2:1

Therefore, the ratio of jerome's share to kevin's share to seth's share is 3:2:1.

8 0
3 years ago
Meg has 26 toy ponies she gets 6 more for her birthday she displays half of them on a shelf and puts the rest in a box how many
otez555 [7]
Because Meg gets an additional 6 ponies, she adds them to her existing 26 ponies. This makes her have 32 ponies.
(CHECK: 26+6=32, 32-6=26)

Since she displays half of her 32, that means 16 of her ponies are on display.
(CHECK: 32/2=16, 16+16=32)

Hopefully this helps!! :)
8 0
3 years ago
Read 2 more answers
Let f = {(x, y) ∈ R × R | xy = 2x + y + 1} There is at least one x ∈ R with no y ∈ R. Find all such problematic x ∈ R. Then rest
loris [4]

Answer:

Problematic x is x = 1

Step-by-step explanation:

Equation:

xy= 2x + y + 1

xy - y = 2x + 1

y(x-1) = 2x + 1

y = (2x+1)/(x-1)

The problematic x is such that when the denominator of the function is 0

x - 1 = 0

x = 1 (the problematic x)

So the domain of f is: x is the subset of R (real number) with the exception of x =/ 1 (x not equal to 1)

To prove this, we can plot the graph and in the graph we can see that as the value of x approaches from negative values to 1, y value will approaches negative infinity, and as the value of x approaches from large positive numbers, y value approaches infinity.

In other words, we'll see an assymptote at x=1

To prove that it is a function, we can do vertical line test by drawing vertical lines accross the graph. We'll see that each line crosses the equation line once hence proving the equation as a function

5 0
3 years ago
Find the value of x in this figure. <br><br> 105<br> 120<br> 110<br> 115
andrezito [222]

Answer:

I believe the answer is 120

4 0
3 years ago
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