Answer:
Step-by-step explanation:
2x+3x=4
5x=4
x=4/5
Answer:
x= 72
Step-by-step explanation:
combine the Xs on one side and then make the denominators the same and solve
8/12X +6 = 3/4X
sub 6 on both sides
then sub 3/4 (aka. 9/12)
-1/12X= -6
multiply by reciprocal (the fraction flip-flopped)(-12) on both sides
and you get 72
Complete question :
Point K on the number line shows Kelvin's score after the first round of a quiz: A number line is shown from negative 10 to 0 to positive 10. There are increments of 1 on either side of the number line. The even numbers are labeled on either side of the number line. Point K is shown on 3. In round 2, he lost 9 points. Which expression shows how many total points he has at the end of round 2? 3 + (−6) = −9, because −9 is 6 units to the left of 3 3 + 6 = −9, because −9 is 6 units to the left of 3 3 + (−9) = −6, because −6 is 9 units to the left of 3 3 + 9 = −6, because −6 is 9 units to the left of 3
Answer: 3 + (−9) = −6, because −6 is 9 units to the left of 3
Step-by-step explanation:
Given the following :
Width of number line = - 10 to + 10
Point K on the number line = +3 ( kelvin's score after the first round of quiz).
If Kelvin losses 9 points in round 2 = - 9
Hence at the end of round 2, He'll have a total of :
Point at the end of round 1 + point lost in round 2
3 + (-9) = - 6
Step-by-step explanation :
Dado que Conner y Jana se multiplican (3568)(39610). El trabajo de Conner El trabajo de Jana (3568)(39610) = 35 + 968 + 10 = 314618 (3568)(39610) = 35⋅968⋅10 = 345680 ¿Es correcto cualquiera de ellos?
Jana y Conner están multiplicando dos números dados. Así que la pregunta es
3568 x 39610
0000
3568x
21408xx
32112xxx
10704xxxx
--------------------------------------
141328480
Así que la respuesta correcta es 141328480.
También los números se pueden agrupar como
3568 = 3000 + 500 + 60 + 8
39610 = 30000 + 9000 + 600 + 10
Después de agrupar multiplicando ambos términos obtenemos la respuesta requerida.
Así que ninguno de los dos tiene la respuesta correcta.
V = 126 if that's what you're looking for