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Brut [27]
3 years ago
10

Can you come up with denominators that contain variables and aren't equal that would require you to change only one denominator

explain please
Mathematics
1 answer:
natali 33 [55]3 years ago
5 0
I think this is what you want.

5 and 10. This would work because 5 goes into 10.

Example: 5, 10, 15, 20
10, 20

They both have a 10
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19. Suppose the diameter of a circle is 10 inches long and a chord is 6 inches long, Find the distance between the chord and the
mihalych1998 [28]

Answer:

10 m

Step-by-step explanation:

x2 + 242 = 262

x2 + 576 = 676

x2 = 100

x = 10

The chord is 10 m from the center of the circle.

3 0
3 years ago
find the sum and product of the zeros of the quadratic polynomial are -1/2 and -3.what is the quadratic equation​
11111nata11111 [884]

Answer:

Step-by-step explanation:

let sum of zeros=s

and product of zeros=p

then quadratic equation is x²-sx+p=0

x²-(-1/2)x+(-3)=0

x²+1/2 x-3=0

2x²+x-6=0

5 0
3 years ago
Find the value of the variable.
MatroZZZ [7]

Answer:

sin 30=14/x

x=14*2

x=28

and

x=√{28²-14²}

x=14√3

7 0
3 years ago
A stack of 30 science flashcards includes a review card for each of the following: 10 insects, 8 trees, 8 flowers, and 4 birds.
coldgirl [10]
All you have to do is make the fraction of it which is 4/30. Then simplify to 2/15 which is your answer.
4 0
2 years ago
What’s the answer?and how do you get it
Kay [80]

the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.

from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.

the standing up sides are simply rectangles of 8x3.

if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.

\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312

7 0
4 years ago
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