A×4 = E, so A = E/4
B/4 = E, so B = 4E
C+4 = E, so C = E=4
D-4 = E, so D = E+4
and
A+B+C+D=100.
now substitute the values that we got of A, B, C and D in the equation, to get,
(E/4) + 4E + E-4 + E+4 = 100
(E/4) + 4E + 2E = 100
(E/4) + 6E = 100
(E+24E)/4 = 100
E+24E = 100×4
25E = 400
E = 400/25
E = 16
i answered same question before too ....
<span>Hope it helps !!!</span>
You should use the website math papa it’ll give you the answer you needed
Let 1st integer = xLet 2nd integer = x + 1 We set up an equation. x(x + 1) = 195 x2 + x = 195 x2 + x - 195 = 0
We will use the quadratic formula: x = (-b ± √(b2 - 4ac) / (2a) x = (-1 ± √(1 - 4(-195))) / 2 x = (-1 ± √(781)) / 2 x = (-1 ± 27.95) / 2 x = 13.48x = -14.78
<span>We determine which value of x when substituted gives us a product of 195.</span> 13.48(14.48) = 195.19-14.48(-13.48) = 195.19 <span>The solution is 2 sets of two consecutive number</span> <span>Set 1</span> The 1st consecutive integer is 13.48The 2nd consecutive integer is 14.48
<span>Set 2</span> The 1st consecutive integer is -14.48The 2nd consecutive integer is -13.48Hopefully this helped, hard work lol :)
