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Illusion [34]
3 years ago
14

Conjecture #2- The product of two rational numbers is always?

Mathematics
1 answer:
Gala2k [10]3 years ago
5 0

Answer:

Rational

Step-by-step explanation:

It is always going to be RATIONAL.

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It is the equivalent dose at 10 mm below a specified point of the body. H p (0.07) is the shallow dose equivalent, also referred to as skin dose equivalent or surface dose equivalent. It is the equivalent dose at 0.07 mm below the surface of the body. Currently, RadTarge II measures Hp (10).

Step-by-step explanation:

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-9, -2, -1.0001, -6, -1.1, -4, and -1.01

Step-by-step explanation:

Anything to the left of -1 on the line means that the number is less than -1.

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A group of young businesswomen wish to open a high fashion boutique in a vacant store, but only if the average income of househo
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<u>Based on the results the boutique could be located there.</u>

<u>Step-by-step explanation:</u>

Note: The average income (the mean) is calculated by summing the total incomes then divided by the total sample–9.

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Based on the results of the p-value, using the 5% significant level <em>we fail to reject the null hypothesis</em> (ú≥25,000).

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Which of the following represents the prime factorization of 72?
Free_Kalibri [48]

Answer:

2^3.3^2

Step-by-step explanation:

{2}^{3 ?}  = 8 \\  {?3}^{2?}  = 9 \\ 8 \times 9 = 72

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Kryger [21]

Answer:

1)

Given the triangle RST with Coordinates  R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.  

Since, the scale factor \frac{5}{3} is greater than 1, the image is enlargement or a stretch.  

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor \frac{5}{3}

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image S{}' are same.

Now, the distances from the center of the dilation at point S to the other points R and T.  

The dilation image will be\frac{5}{3} of each of these distances,

SR=3, so S{}'R{}'=5 ;


ST=3, so S{}'T{}'=5  

Now, draw the image of RST i.e R'S'T'

Since, RT=3\sqrt{2} [By using hypotenuse of right angle triangle] and R{}'T{}'=5\sqrt{2}.


2)

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of \overline{AC} i.e \overline{AB}\cong \overline{BC}

In the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

Since, there is no included angle in these triangles.

∴ \Delta ABD is not congruent to \Delta CBD .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that  \overline{AD}\cong \overline{CD}.

therefore, in the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{AD}\cong \overline{CD}   (SIDE)           [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

therefore by, SSS postulates \Delta ABD\cong \Delta CBD.

3)

Given that:  \angle1=\angle 3 are vertical angles, as they are formed by intersecting lines.

Therefore

, by the definition of linear pairs

\angle 1 and \angle 2 and \angle 3  and \angle 2 are linear pair.

By linear pair theorem, \angle 1 and \angle 2   are supplementary, \angle 2 and \angle 3  are supplementary.

m\angle1+m\angle 2=180^{\circ}

m\angle2+m\angle 3=180^{\circ}

Equate the above expressions:

m\angle 1+m\angle 2=m\angle 2+m\angle 3

Subtract the angle 2 from both sides in the above expressions

∴m\angle 1=m\angle 3

By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.


therefore, \angle 1\cong \angle 3.















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2 years ago
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