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rusak2 [61]
3 years ago
10

Please help me out !

Mathematics
2 answers:
hjlf3 years ago
7 0

Answer:

a) \frac{-9}{-10}       b) \frac{-3}{5} , -\frac{3}{5}

Step-by-step explanation:

a) Since the fraction you are dealing with, \frac{9}{10} , is positive, we will need to find a fraction that has either two negative numbers or two positive numbers. The first option, -\frac{9}{10} , the entire fraction is negative, which does not match the given fraction. The second option, \frac{-9}{10} , only one number is negative, and that does not match either. The third option, \frac{9}{-10} , is the same as the second option, only one negative number. However the last option, \frac{-9}{-10} , has both a negative number in the numerator and denominator which both cancel out to \frac{9}{10}, making the fraction positive.

b) The fraction we are dealing with here, \frac{-3}{5} , is negative, so we need to find a fraction that has only one negative number. The first option, \frac{3}{5} , has no negative numbers, making it positive, which does not match. The second option, \frac{3}{-5} , has one negative number, which matches our fraction. The third fraction, \frac{-3}{-5} has two negative numbers making the fraction \frac{3}{5}, positive, which does not match. Then the last option, -\frac{3}{5} the whole fraction is negative which matches our fraction.

Ahat [919]3 years ago
3 0

Answer:

a) -9/-10 (4th)

b) -3/5 (4th)

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What is a solution to the inequality y&gt;3x-9
Alchen [17]

Answer:

Many answers including (0,0)

Step-by-step explanation:

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7 0
3 years ago
Given ACM, angle C=90º. AP=9, PM=12. Find AC, CM, AM.
Gnesinka [82]

Answer:

AM = 25, AC = 15, CM = 20

Step-by-step explanation:

The given parameters are;

In ΔACM, ∠C = 90°, \overline{CP} ⊥ \overline{AM}, AP = 9, and PM = 16

\overline{AC}² + \overline{CM}² = \overline{AM}²

\overline{AM} = \overline{AP} + PM = 9 + 16 = 25

\overline{AM} = 25

\overline{AC}² = \overline{AP}² + \overline{CP}² = 9² +  \overline{CP}²

∴ \overline{AC}² = 9² +  \overline{CP}²

Similarly we get;

\overline{CM}² = 16² + \overline{CP}²

Therefore, we get;

\overline{AC}² + \overline{CM}² = 9² +  \overline{CP}² + 16² + \overline{CP}² = \overline{AM}² = 25²

2·\overline{CP}² = 25² - (9² + 16²) = 288

\overline{CP}² = 288/2 = 144

\overline{CP} = √144 = 12

From \overline{AC}² = 9² +  \overline{CP}², we get

\overline{AC} = √(9² +  12²) = 15

\overline{AC} = 15

From, \overline{CM}² = 16² + \overline{CP}², we get;

\overline{CM} = √(16² + 12²) = 20

\overline{CM} = 20.

3 0
3 years ago
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