1. Solve: RGTG = 171; RRTR = 171; RR = 3RG ; TR = TG – 6

2. Find three consecutive even integers such that 4 times the product of the first and the third is 28

3.greater than the product of -10 and the sum of the second and the third.

Solve by completing the square: 3x2 + 8 = 5x

4.Evaluate: 567.3 x10^28  1232.3 10^28

5.evaluate:(3.42)^-2.34

6.Find the equation of the line that passes through (-8, 2) and (5, -7)

7.find c= xp=m(b/1+c + a/p)

 

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1 Answer

1. Solve: RGT= 171; RRT= 171; R= 3R; T= T– 6

RRT=171=3RG(T– 6)=3RGTG– 18RG=3×171– 18RG, 18RG=342, RG=342/18=19.

T=171/19=9, T=9-6=3, RR=171/3=57.

SOLUTION: RG=19, T=9, RR=57, T=3.

2. Find three consecutive even integers such that 4 times the product of the first and the third is 28. Let integers be 2(n-1), 2n, 2(n+1).

4(4(n2-1))=28, 4(n2-1)=7. No integer or rational solutions. If the product is 48 instead of 28:

16(n2-1)=48, n2-1=3, n2=4, n=2 or -2, so the integers would be 2, 4, 6 or -6, -4, -2. If the product is 128 instead of 28:

n2-1=128/16=8 and n=±3, and the integers would be 4, 6, 8 or -8, -6, -4.

3.greater than the product of -10 and the sum of the second and the third. Incomplete question.

Solve by completing the square: 3x2 + 8 = 5x, x2-5x/3=-8/3, x2-5x/3+25/36=25/36-8/3, (x-5/6)2=-71/36. No real solution. Complex solution x=5/6±i√71/6.

Perhaps the quadratic equation should have been 3x2+8x=5, then x2+8x/3+16/9=5/3+16/9, (x+4/3)2=31/9, x=-4/3±√31/3 or (-4±√31)/3.

Or it could have been 3x2+8x=-5, then x2+8x/3+16/9=-5/3+16/9, (x+4/3)2=1/9, x=-4/3±⅓, that is, -5/3 or -1.

4.Evaluate: 567.3 x1028  1232.3 1028=567.3/1232.3=0.46036 approx or 5673/12323.

5.evaluate:(3.42)-2.34=0.05628 approx.

6.Find the equation of the line that passes through (-8, 2) and (5, -7)

Slope=(2+7)/(-8-5)=-9/13, y-2=-9(x+8)/13, y=-9x/13-72/13+2=-9x/13-46/13.

7.find c= xp=m(b/1+c + a/p): xp=m(b/1+c + a/p)=m(bp+a+ac)/(p+cp), xp(p+cp)=mbp+am+mac, xp2+cxp2=mbp+am+mac,

c(xp2-ma)=mbp+am-xp2, c=(mbp+am-xp2)/(xp2-ma).

by Top Rated User (1.2m points)

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