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ivann1987 [24]
3 years ago
9

Is 11/13 a ratio in simplest form

Mathematics
1 answer:
Molodets [167]3 years ago
5 0
Yes 11/13 cannot be simplified anymore
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Consider the decimal 0.101001000100001000.... do you think this decimal represents a rational number?why or why not
Katarina [22]
The decimal is irrational because it is not repeating.
4 0
3 years ago
Points J and K lie in plane H.
Vlad [161]

Answer: 1

Step-by-step explanation:

From the given picture, it can be seen that there is a plane H on which two pints J and K are located.

One of the Axiom in Euclid's geometry says that <em>"Through any given two points X and Y, only one and only one  line can be drawn "</em>

Therefore by Axiom in Euclid's geometry , for the given points J and K in plane H , only one line can be drawn  through points J and K.

6 0
3 years ago
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(NEED IT FAST) Clarke makes 27 party favors in 4 hours. Carolynn makes 35 party favors in five hours. Who is faster? What is the
ss7ja [257]

Answer: Carolynn is faster.

Step-by-step explanation: Clarke makes 6.75 favors/hour, while Carolynn makes 7 favors/hour. (7 > 6.75)

7 0
3 years ago
Johanna can jog 24 feet in 5 seconds. If she jogs at the same rate, how many feet can she jog in 8 seconds? PLSS help!
Fynjy0 [20]
38.4? Since if she runs 24 feet in five seconds add that two times she would run 48 feet in ten second so.. divide 24 with 5 and get 4.8. So 48 minus 9.6 makes 38.4.
6 0
3 years ago
How do you solve 2x^2 - 7x -5= 0 in quadratic form?
lesantik [10]

Answer:

Once the equation is in standard form, factor the quadratic expression. 2x2 + 7x + 3 = 0 (2x + 1)(x + 3) = 0. Using the Zero Product Property set ...

2x2 + 7x = -3

2x2 + 7x + 3 = 0

Once the equation is in standard form, factor the quadratic expression.

2x2 + 7x + 3 = 0

(2x + 1)(x + 3) = 0

Using the Zero Product Property set each factor equal to 0 and solve for x.

2x + 1 = 0  

2x + 1 - 1 = 0 - 1 x + 3 = 0

2x = -1 x + 3 - 3 = 0 - 3

2x 2 = -1 2 x = -3

x = -1 2  

The solutions to the equation are -1 2 and -3.

7 0
3 years ago
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