Nadiya determines that there are no fractions equivalent to 2/10 with a denominator greater than 10, but less than 20, that have
a whole number numerator and denominator. Is Nadiya correct?
Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to 2/10 is 4/20.
Yes, Nadiya is correct. By multiplying the whole fraction by any whole number, each equivalent fraction will continue to have a denominator of 10.
No, Nadiya is incorrect. She could multiply both the numerator and denominator of the unit fraction equivalent to 2/10 by 3 and arrive at the equivalent fraction 3/15.
No, Nadiya is incorrect. She could add any number between 1 and 8 to the numerator and denominator and arrive at an equivalent fraction with a denominator greater than 10 but less than 20.
Yes, Nadiya is correct. By multiplying the numerator and denominator by 2, the first fraction equivalent to 2/10 is 4/20.
Yes, Nadiya is correct. By multiplying the whole fraction by any whole number, each equivalent fraction will continue to have a denominator of 10.
No, Nadiya is incorrect. She could multiply both the numerator and denominator of the unit fraction equivalent to 2/10 by 3 and arrive at the equivalent fraction 3/15.
No, Nadiya is incorrect. She could add any number between 1 and 8 to the numerator and denominator and arrive at an equivalent fraction with a denominator greater than 10 but less than 20.
2 answers:
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Answer:
D
Step-by-step explanation:
The line x originates from the center of the circle and forms a perpendicular line with line RS, which means that the line is bisected.
That means that line RT is half of line RS, or 7.
Then we can use the Pythagorean Theorem to find the measure of x.
b = √32
b ≈ 5.66
Answer:
<u><em>(-1,1)</em></u>
Step-by-step explanation:
We can solve this by either graphing and finding ther point the lines intersect, or mathematically, I'll do both.
<u>Graphing:</u>
<u>Mathematically:</u>
−2x + 4y = 6
y = 2x + 3
See the attached graph. The lines intersect at (-1,1)
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I'll rearrange the first equation (to make it easier for me):
−2x + 4y = 6
4y = 2x + 6
y = (1/2)x + 1.5
Now lets substitute the second equation into the first so that we can eliminate y:
y = 2x + 3
[(1/2)x + 1.5] = 2x + 3
- (3/2)x = (3/2)
x = -1
If x = -1:
y = 2(-1) + 3
y = 1
The solution is x = -1 and y = 1, or (-1,1)
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Both approaches give us (-1,1), the solution to the system of equations. It is the only point that satisfies both equations.
