Find the of-51- -203- -36 and -28

Question

Toppr Admin · Accepted Answer

-51-203-36-28-51-203-36-28-Here- we will -1-st- solve square bracket-therefore -51-203-36-28-51-203-36-28-254-8-246-246- is the sum of -51- -203- -36- and -28-

Find the sum of
$$-51, \,-203, \,36$$ and $$-28$$

$$-51+(-203)+36+(-28)$$
$$=[-51+(-203)]+[36+(-28)]$$
Here, we will $$1^{st}$$ solve square bracket
$$\therefore [-51+(-203)]+[36+(-28)]=(-51-203)+(36-28)$$
$$=-254+8$$
$$=-246$$
$$-246$$ is the sum of $$-51, \,-203, \,36$$ and $$-28$$

Find the sum of$$-51, \,-203, \,36$$ and $$-28$$

$$-51+(-203)+36+(-28)$$$$=[-51+(-203)]+[36+(-28)]$$Here, we will $$1^{st}$$ solve square bracket$$\therefore [-51+(-203)]+[36+(-28)]=(-51-203)+(36-28)$$$$=-254+8$$$$=-246$$$$-246$$ is the sum of $$-51, \,-203, \,36$$ and $$-28$$

Find the sum of
$$-51, \,-203, \,36$$ and $$-28$$

$$-51+(-203)+36+(-28)$$
$$=[-51+(-203)]+[36+(-28)]$$
Here, we will $$1^{st}$$ solve square bracket
$$\therefore [-51+(-203)]+[36+(-28)]=(-51-203)+(36-28)$$
$$=-254+8$$
$$=-246$$
$$-246$$ is the sum of $$-51, \,-203, \,36$$ and $$-28$$