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Let's lay out the known information in equations, where B is the number of boys and G is the number of girls. Then we express the number B in terms of G, which is twice as much less 12, or (2G-12) B + G = 51 , and B = 2G - 12 Now, replace the B in the first equation with its value from the second equation (2G - 12) + G = 51 Remove the brackets 2G - 12 + G = 51 Add 12 to both sides 2G + G = 51 + 12 3G = 63 Divide both sides by 3 G = 21 Now, to check if we're right, Replace G in the first equation B + 21 = 51 B = 30 If that's right then 30 = 2(21) - 12 30 = 42 - 12 And it is, so we know that there are 21 girls and 30 boys in ninth grade.
One fourth of 8 is 2.
Points: (-12, -17) and (21, 5) Slope: 2/3
hmm... i don' think you have the completed the question. This one is pretty hard! Solving quadratics can be difficult and ompleting the square is one method for solving a quadratic equation.Is the equation -84x2-8x-12=0 or -84x2+8x - 12 = 0 or -84x2 = 8x - 12?If it is the first one:Factor out the 4:-4 (21x2 + 2x + 3) = 021x2 + 2x + 3 = 0 (divide by 21 each term to both sides)x2 + (2/21)x + 3/21 = 0 (subtract 3/21 to both sides)x2 + (2/21)x = -3/21 (add (1/2 of 2/21)2 or (1/21)2 to both sides)x2 + (2/21)x + (1/21)2 = -3/21 + 1/212(x + 1/21)2 = -63/212 + 1/212(x + 1/21)2 = -62/212 (square root to both sides)x + 1/21 = ± √-62/212x + 1/21 = ± (i/21)√62 (subtract 1/21 to both sides)x = -1/21 ± (i/21)√62x = -1/21 (1 - i√62) orx = -1/21 (1+ i√62)If it is the third one:-84x2 = 8x - 12 (divide by -84 to both sides)x2 = -(2/21)x + 3/21 (add (2/21)x to both sides)x2 + (2/21)x = 3/21 (add (1/2 of 2/21)2 or (1/21)2 to both sides)x2 + (2/21)x + (1/21)2 = 3/21 + 1/212(x + 1/21)2 = 63/212 + 1/212(x + 1/21)2 = 64/212 (square root to both sides)x + 1/21 = ± √64/212x + 1/21 = ± 8/21 (subtract 1/21 to both sides)x = -1/21 ± 8/21x = -1/21 + 8/21 = 7/21 = 1/3 orx = -1/21 - 8/21 = -9/21 = -3/7
It is: 2 times the square root of 21 and it's an irrational number