I believe it is 56, can someone confirm this?
C 5z+11-4p.
5(z+3)-4(1+p)
5z+15-4-4p
5z+11-4p
The answer is
<span>x+1</span>, <span>x+2</span>, <span>x+3</span>, <span>x+4</span> and <span>x+5</span>.
The sum of these six integers is 393 so we can write:
<span>x+x+1+x+2+x+3+x+4+x+5=393</span>
<span>6x+1+2+3+4+5=393</span>
<span>6x+15=393</span>
<span>6x+15−15=393−15</span>
<span>6x+0=378</span>
<span><span><span>6x</span>6</span>=<span>3786</span></span>
<span>x=63</span>
Because the first integer is 63 then the third would be <span>x+2</span> or <span>63+2=<span>65</span></span>
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<span><span>Hope its helps :)</span></span>
5.96.....the last digit, 6, is in the hundredths place...
5 96/100 or 596/100 which reduces to 149/25