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Butoxors [25]
3 years ago
12

Please help with 12 through 14

Mathematics
1 answer:
BabaBlast [244]3 years ago
4 0
The formula to find the area of a circle is A=r^2pie
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The height of dachshunds is usually 1/3 their length. If Mollie is 20 inches long, how tall is she?
arlik [135]

This is a proportion problem.

Let's make her Length x and her height 1/3 x.

1/3 x = 20 inches

x = 6 2/3 inches

Molly's height is 6 2/3 inches.

Check it.

6 2/3 x 3 = ? (change the mixed number to a fraction)

20/3 x 3 = ? (multiply)

60/3 = ? (divide)

20 inches. Yes, it checks.

8 0
3 years ago
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Factorizatioun of X^2+5x-6
NNADVOKAT [17]

x² + 5x - 6 can be expressed as (x + 6)(x - 1).

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3 years ago
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0.009 ia 1/10 of what number?
d1i1m1o1n [39]
0.009 x 10 = 0.09

it may not be correct, but here you go.
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3 years ago
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I need some heckin help cuz i suck at math :)
vladimir1956 [14]

Answer:

Step-by-step explanation:

Reasons

2. If two lines are parallel, their corresponding angles are congruent.

3. Congruent angles are equal... Why do they even have this step?

4. A straight line forms a linear pair.

5. Angles in a linear pair are supplementary.

6. ∠1 is supplementary to ∠3. Reason: If an angle is congruent to an angle that is supplementary to a third angle, the first and third angles are congruent OR Transitive Property.

7 0
3 years ago
Solve this query plzzz​
Olin [163]

Step-by-step explanation:

\frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   = \tan \:A \\  \\ LHS =  \frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   \\  \\  =  \frac{2\sin \:2A.\cos \:2A}{\cos \:2A}   \times  \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\  \\  = 2\sin \:2A   \times  \frac{1 - 2 { \cos}^{2}A  +  1}{1 - 2 { \cos}^{2}2A  + 1 } \\  \\  = 2\sin \:2A   \times  \frac{2- 2 { \cos}^{2}A  }{2 - 2 { \cos}^{2}2A   } \\  \\   = 2\sin \:2A   \times  \frac{2(1 - { \cos}^{2}A)  }{2 (1-  { \cos}^{2}2A)   } \\  \\   = 2\sin \:2A   \times  \frac{1 - { \cos}^{2}A}{1-  { \cos}^{2}2A   } \\  \\     = 2\sin \:2A   \times  \frac{ { \sin}^{2}A}{{ \sin}^{2}2A   } \\  \\    = 2  \times  \frac{ { \sin}^{2}A}{{ \sin}2A   } \\  \\   = 2  \times  \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \:   A } \\  \\   = \frac{ { \sin}A}{ \cos \:   A }  \\  \\  = tan \: A \\  \\  = RHS \\

7 0
4 years ago
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