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dedylja [7]
3 years ago
6

Which statement that describes show that figure a is congruent to figure b

Mathematics
1 answer:
Masja [62]3 years ago
7 0

Answer:

E.

Step-by-step explanation:

Translation: <u>s</u><u>l</u><u>i</u><u>d</u><u>e</u><u> </u>of the shape or figure, it does NOT change its shape, size, and form, count how many units the shape moves.

Reflection: <u>f</u><u>l</u><u>i</u><u>p</u> of the shape, does NOT change its shape, size, and form, either can be flipped or reflected over the x or y axis.

Rotation: <u>t</u><u>u</u><u>r</u><u>n</u> of the shape, does NOT change its shape, size, and form, either can be rotated counter clock wise left, (opposite of a clock direction), or clockwise, the direction a clock moves right. It can also be rotated 90° (1 turn), 180° (2 turns), 270° (3 turns) or 360° (4 turns). In highschool, you will get introduced to more degrees of rotation, this is just middle school level.

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Which value of www makes 14=11+\dfrac{w}{8}\cdot614=11+ 8 w ​ ⋅614, equals, 11, plus, start fraction, w, divided by, 8, end frac
Liono4ka [1.6K]

Answer:

Option B) w = 4

Step-by-step explanation:

We have to find the value of w to make the given expression true.

The given expression is:

14=11+\dfrac{w}{8}\cdot6\\\\14-11 = \dfrac{w}{8}\cdot6\\\\3 = \dfrac{w}{8}\cdot6\\\\\Rightarrow w = \dfrac{3 \cdot 8}{6} = 4

Option B) w = 4 is the correct answer.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Recall, we have five connectives in propositional logic ¬, ∧, ∨, →, ↔ (negation, conjunction, disjunction, conditional and bicon
Free_Kalibri [48]

Answer:

(a) ¬(p→¬q)

(b) ¬p→q

(c) ¬((p→q)→¬(q→p))

Step-by-step explanation

taking into account the truth table for the conditional connective:

<u>p | q | p→q </u>

T | T |   T    

T | F |   F    

F | T |   T    

F | F |   T    

(a) and (b) can be seen from truth tables:

for (a) <u>p∧q</u>:

<u>p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q</u>

T | T |  F  |   F     |    T       |  T

T | F |  T  |  T      |    F       |  F

F | T |  F  |  T      |    F       |  F

F | F |  T  |  T      |    F       |  F

As they have the same truth table, they are equivalent.

In a similar manner, for (b) p∨q:

<u>p | q | ¬p | ¬p→q | p∨q</u>

T | T |  F  |   T     |    T    

T | F |  F  |   T     |    T    

F | T |  T  |   T     |    T    

F | F |  T  |   F     |    F    

again, the truth tables are the same.

For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))

4 0
3 years ago
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