Q: If $$A$$ < $$0.5$$ < $$B$$ < $$15,$$ and $$30$$ < $$C$$ < $$60,$$ what is the relative order of the reciprocals $$\frac{1}{A},\frac{1}{B},\frac{1}{C}?$$

Since $$A$$ $$2.$$Since $$0.5$$<$$B$$<$$15,$$ we have$$\frac{1}{15}$$<$$\frac{1}{B}$$<$$2.$$So$$0.066\ldots$$<$$\frac{1}{B}$$<$$2.$$Since $$30$$<$$C$$<$$60,$$ we have$$\frac{1}{60}$$<$$\frac{1}{C}$$<$$\frac{1}{30}.$$So$$0.016\ldots$$<$$\frac{1}{C}$$<$$0.033\ldots.$$Thus the ordering from smallest to largest is:$$\frac{1}{C}$$<$$\frac{1}{B}$$<$$\frac{1}{A}.$$Final answer: $$\frac{1}{C}$$<$$\frac{1}{B}$$<$$\frac{1}{A}$$

Q: If $$3x-y=9$$ and $$\frac{x}{y}=2$$, then $$x=$$?

From $$\frac{x}{y}=2$$ we get $$x=2y$$Substitute into $$3x-y=9$$$$3(2y)-y=9$$$$6y-y=9$$$$5y=9$$$$y=\frac{9}{5}$$Then$$x=2\cdot\frac{9}{5}=\frac{18}{5}$$$$\frac{18}{5}=3.6$$Final answer: $$3.6$$

Q: If $$\frac{1}{x}=0.25$$, then $$\frac{1}{x+3}=$$?

Rewrite$$0.25=\frac{1}{4}$$So$$\frac{1}{x}=\frac{1}{4}$$Thus$$x=4$$Now compute$$\frac{1}{x+3}=\frac{1}{4+3}=\frac{1}{7}$$Final answer: $$\frac{1}{7}$$

Q: If $$3x-2y=7$$ and $$4x+y=5$$, then $$x=$$

Use $$4x+y=5$$So$$y=5-4x$$Substitute into $$3x-2y=7$$$$3x-2(5-4x)=7$$$$3x-10+8x=7$$$$11x=17$$$$x=\frac{17}{11}$$Final answer: $$\frac{17}{11}$$

Q: If $$3x-2y=8$$, then $$6y-9x=$$

Start with $$3x-2y=8$$Multiply both sides by:$$-3$$ $$-9x+6y=-24$$Rewrite.$$6y-9x=-24$$Final answer: $$-24$$

Question 1

$$6x-3y=9$$ $$6x+3y=21$$ Find the value of $$|x|+|y|$$.

Accepted Answer

Add the equations.$$12x=30$$$$x=\frac{30}{12}=\frac{5}{2}$$Substitute into the first equation.$$6\left(\frac{5}{2}\right)-3y=9$$$$15-3y=9$$$$-3y=-6$$$$y=2$$Compute.$$|x|=\frac{5}{2}$$$$|y|=2$$$$|x|+|y|=\frac{5}{2}+2=\frac{9}{2}$$$$\frac{9}{2}=4.5$$Final answer: $$4.5$$

Question 2

Solve the inequality. $$5x-7$$ > $$3-2x$$ Select all values of $$x$$ that satisfy the inequality.

Accepted Answer

Solve the inequality.$$5x-7$$>$$3-2x$$Add $$2x$$.$$7x-7$$>$$3$$Add $$7$$.$$7x$$>$$10$$$$x$$>$$\frac{10}{7}$$Check choices. $$2,3,4,5$$ satisfy.Final answer: $$2,3,4,5$$

Question 3

If $$A$$ < $$0.5$$ < $$B$$ < $$15,$$ and $$30$$ < $$C$$ < $$60,$$ what is the relative order of the reciprocals $$\frac{1}{A},\frac{1}{B},\frac{1}{C}?$$

Accepted Answer

Since $$A$$<$$0.5,$$ we have$$\frac{1}{A}$$>$$2.$$Since $$0.5$$<$$B$$<$$15,$$ we have$$\frac{1}{15}$$<$$\frac{1}{B}$$<$$2.$$So$$0.066\ldots$$<$$\frac{1}{B}$$<$$2.$$Since $$30$$<$$C$$<$$60,$$ we have$$\frac{1}{60}$$<$$\frac{1}{C}$$<$$\frac{1}{30}.$$So$$0.016\ldots$$<$$\frac{1}{C}$$<$$0.033\ldots.$$Thus the ordering from smallest to largest is:$$\frac{1}{C}$$<$$\frac{1}{B}$$<$$\frac{1}{A}.$$Final answer: $$\frac{1}{C}$$<$$\frac{1}{B}$$<$$\frac{1}{A}$$

Question 4

If $$3x-y=9$$ and $$\frac{x}{y}=2$$, then $$x=$$?

Accepted Answer

From $$\frac{x}{y}=2$$ we get $$x=2y$$Substitute into $$3x-y=9$$$$3(2y)-y=9$$$$6y-y=9$$$$5y=9$$$$y=\frac{9}{5}$$Then$$x=2\cdot\frac{9}{5}=\frac{18}{5}$$$$\frac{18}{5}=3.6$$Final answer: $$3.6$$

Question 5

If $$\frac{1}{x}=0.25$$, then $$\frac{1}{x+3}=$$?

Accepted Answer

Rewrite$$0.25=\frac{1}{4}$$So$$\frac{1}{x}=\frac{1}{4}$$Thus$$x=4$$Now compute$$\frac{1}{x+3}=\frac{1}{4+3}=\frac{1}{7}$$Final answer: $$\frac{1}{7}$$

Question 6

If $$3x-2y=7$$ and $$4x+y=5$$, then $$x=$$

Accepted Answer

Use $$4x+y=5$$So$$y=5-4x$$Substitute into $$3x-2y=7$$$$3x-2(5-4x)=7$$$$3x-10+8x=7$$$$11x=17$$$$x=\frac{17}{11}$$Final answer: $$\frac{17}{11}$$

Question 7

If $$3x-2y=8$$, then $$6y-9x=$$

Accepted Answer

Start with $$3x-2y=8$$Multiply both sides by:$$-3$$ $$-9x+6y=-24$$Rewrite.$$6y-9x=-24$$Final answer: $$-24$$

Question 8

If $$y-2x$$ > $$10$$ and $$x-y$$ > $$20$$, which of the following are possible values of $$x$$? Indicate all such values.

Accepted Answer

Start with the inequalities.$$y-2x$$>$$10$$$$y$$>$$2x+10.$$$$x-y$$>$$20$$$$y$$<$$x-20.$$Combine:$$2x+10$$<$$y$$<$$x-20.$$For this to be possible, we need:$$2x+10$$<$$x-20.$$Solve:$$2x+10$$<$$x-20$$$$x+10$$<$$-20$$$$x$$<$$-30.$$So all possible values of $$x$$ must satisfy:$$x$$<$$-30.$$Final answer: $$x$$<$$-30$$

Quiz 1

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