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Galina-37 [17]
3 years ago
10

(-8x + 1) – (8x – 1) = ??

Mathematics
1 answer:
Bad White [126]3 years ago
6 0

Answer:0

Step-by-step explanation:

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Topic is Sets, please help​
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I couldn’t see the whole picture so I’m not sure if I could really help I’m sorry maybe next time
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15.5 × [(2 × 2.4) + 3.2] – 16
ch4aika [34]

Answer:

108

Step-by-step explanation:

5 0
3 years ago
HEEEEEELP !
iogann1982 [59]

If you meant to write y=23x-4 then none of the lines are perpendicular.

I suspect you intended y=2/3x-4, then line D, y=-3/2x-4 is perpendicular.

For any line y=mx+b, you have to "invert" m to get  a perpendicular line. Inverting in this case means: flip numerator and denominator and add a minus sign.

So 2/3 becomes -3/2, hence answer D.

The 'inverted' m is called the opposite reciprocal. That's your word of the day.


6 0
3 years ago
There are 84 boys in a senior class of 146 students. find the ratio of boys to girls. (note: if you know the total number of stu
vladimir1956 [14]
84 to 62. You have to subtract 146-84 to get the number of girls from the class.
4 0
3 years ago
Please Identify the congruent triangles in the figure.
Tpy6a [65]

The missing steps are each right angles and \angle R \cong \angle O.

Solution:

Step 1: Given data:

\overline {P Q} \cong \overline{M N}

\overline{Q R} \cong \overline{N O}

\overline{P R} \cong \overline{M O}

Step 2: In the two polygons,

\angle Q =90^\circ and \angle N =90^\circ

\angle Q \cong \angle N (Each right angle)

Step 3: Given

\angle P \cong \angle M

Step 4: By third angle theorem,

If two angles in one triangle are congruent to the two angles in the other triangle, then the third angles in the triangles also congruent.

\angle R \cong \angle O

Step 5: By the definition of congruent polygons,

If two same shape polygons have all the angles are congruent and all the corresponding sides are congruent then the polygons are congruent.

Hence \Delta P Q R \cong \Delta M N O.

Therefore the missing steps are each right angles and \angle R \cong \angle O.

5 0
3 years ago
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