Answer: (a) 9
<u>Step-by-step explanation:</u>
1st multiple of 3: 1(3x)
2nd multiple of 3: 2(3x)
3rd multiple of 3: 3(3x)
Sum = 4(1st multiple of 3) + 6
1(3x) + 2(3x) + 3(3x) = 4(3x) + 6
3x + 6x + 9x = 12x + 6
18x = 12x + 6
6x = 6
x = 1
Largest (3rd) multiple is: 3(3x)
= 9x
= 9 · 1
= 9
Answer:
The correct option is B.
Step-by-step explanation:
Given information: AB\parallel DCAB∥DC and BC\parallel ADBC∥AD .
Draw a diagonal AC.
In triangle BCA and DAC,
AC\cong ACAC≅AC (Reflexive Property of Equality)
\angle BAC\cong \angle DCA∠BAC≅∠DCA ( Alternate Interior Angles Theorem)
\angle BCA\cong \angle DAC∠BCA≅∠DAC ( Alternate Interior Angles Theorem)
The ASA (Angle-Side-Angle) postulate states that two triangles are congruent if two corresponding angles and the included side of are congruent.
By ASA postulate,
\triangle BCA\cong \triangle DAC△BCA≅△DAC
Therefore option B is correct
Answer:
1.thursday
his expenses was p26.85
2.monday he spent p40.8
3.18.90+23.15+11.80+16.35+9.20=approximately p55.6×30 days=p1,668
4.693.80-1,668
so he savings is completely enough
When you can divide, or multiply by the same number. for numerator and denominator.
