Choice C
$$a(4-a)-5(a+7)=4a-a^2-5a-35=-a^2-a-35$$
Choice G
$$\frac{1}{4}=0.25 \\$$
$$ \Rightarrow 0.03 \leq 0.2 \leq 0.25 \\$$
$$ \Rightarrow 0.03 \leq 0.2 \leq \frac{1}{4}$$
Choice C
$$x^2=29-4=25 \\$$
$$ \Rightarrow x^2-4=25-4=21$$
Choice H
Since T and U bisect QR and QS, respectively, TU//RS. Thus,
$$\angle TUQ=\angle S=60° \\$$
$$ \Rightarrow \angle TUS=180°-\angle TUQ=180°-60°=120°$$
Choice C
The line is going downwards. Therefore the slope is negative.
Choice K
$$\sqrt{81}=9 \\ $$
$$\sqrt{99}\approx\sqrt{100}=10$$
$$ \\ 9 \leq 9.2371 \leq 10 \\$$
$$ \Rightarrow 81\leq the \ number \leq 99$$
Choice E
$$p=\frac{10-2}{10}=\frac{4}{5}$$
Choice K
$$\frac{x_{A}+(-3)}{2}=1 \\$$
$$ \Rightarrow x_{A}=5 \\$$
$$ \frac{y_{A}+4}{2}=2 \\$$
$$ \Rightarrow y_{A}=0 $$
Choice B
$$2x+8=116 \\ $$
$$\Rightarrow x=54$$
Choice H
$$500+200x=$2,200 \\$$
$$ \Rightarrow x=$8.50$$
Choice D
$$(3x+6)(2x-1)=6x^2-3x+12x-6=6x^2+9x-6$$
Choice J
A box of 25 candies at Tamika's Treat costs $4.25. Thus the money left is $10.00 - $4.25 = $5.75, which is 23 quarters:
$$\frac{$5.75}{$0.25}=23$$
Choice B
$$\frac{$3.75}{20}=$0.1875\approx $0.19$$
Choice F
Only the equation in F fulfils all the given prices and the number of candies given in the table.
Choice E
The opposite of 'at least 1' is 'no one'.
Choice J
Choice E
$$\frac{9\times9\times12}{3\times3\times3}=36$$
Choice J
$$f(-4)=-4(-4)^{3}-4(-4)^{2}
=-4(-64)-4(16)
=256-64
=192$$
Choice A
$$x+2y=4 \\ $$
$$-2x+y=7 \\$$
$$\Rightarrow x=-2, \ y=3$$
Choice G
$$x=\sqrt{8^2+9^2}\approx12 \\$$
$$Perimeter = 8+10+12+19=49\approx50$$
Choice C
$$5\times7-4\times6=11$$
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Choice H
$$\frac{BC}{AB}=\frac{BD}{BE} \\$$
$$ \Rightarrow \frac{BC}{3}=\frac{12}{4} …
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Choice D
When x=4, y can have …
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Choice J
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Choice D
Let the number of times …
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Choice K
Since a and b are …
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Choice C
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Choice H
$$40\%\times250=60\%x \\$$
$$ \Rightarrow x=166 …
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Choice A
$$-2x-6y>2y-4 \\$$
$$ \Rightarrow …
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Choice F
$$ \tan\alpha=\frac{\sin\alpha}{\cos\alpha} \\$$
$$ \Rightarrow …
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Choice B
$$3k+5k+3k+5k=96 \ cm \\ $$ …
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Choice G
$$d^2=\frac{16\times8}{2} \\$$
$$ \Rightarrow d=8 …
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Choice B
$$\frac{3.6}{12}=30 \%$$
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Choice F
$$JD=\sqrt{5^2+12^2}=13$$
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Choice B
$$length \ of arc\ \widehat{CD}=\frac{2\pi …
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Choice H
12 - 3.6 = 8.4
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Choice A
4 - 1 = 3
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Choice J
$$P=\frac{3^ne^{-3}}{n!}=\frac{3^2\times0.05}{2!}\approx0.23$$
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Choice B
The amplitude of f(x)=cos(x) is …
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Choice H
$$\frac{4}{3}\pi x^3=12 \\ \Rightarrow x=\sqrt[3]{\frac{9}{\pi}}$$
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Choice D
The median will be the …
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Choice H
$$\tan40°=\frac{height}{50}\\$$
$$ \Rightarrow height=50\times\tan40°=42$$
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Choice B
Sea Horse:
$$\frac{255-50-50}{0.25}=620$$
Ocean Blue: …
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Choice G
0.05 fits the data best.
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Choice A
$$2x-5y=-5 \\ $$
$$\Rightarrow 2x+5=5y …
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Choice J
$$\frac{x}{2\frac{1}{2}}=\frac{2\frac{1}{4}}{1\frac{1}{2}} \\$$
$$ \Rightarrow x= …
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Choice E
$$\frac{12x^6}{3x^2}-\frac{9x^2}{3x^2}=4x^4-3$$
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Choice D
$$2x=5\times7 \\$$
$$ \Rightarrow x=17.5$$
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Choice F
$$\frac{[5+(5+x)]\cdot x}{2}=12 \\$$
$$ \Rightarrow …
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Choice B
The square of a number …
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Choice K
$$(x-1)(x+2)=x^{2}+x-2$$
$$x^{2}+x-2=x^{2}+(a+2)x+a+b$$
$$\require{cancel}$$
$$\cancel{x^{2}}+x-2=\cancel{x^{2}}+(a+2)x+a+b$$
$$\Rightarrow …
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Choice E
Since the given equation is …
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Choice J
The slopes of two lines …
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Choice A
The shaded area lies below …
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Choice K
$$(\sqrt{3})^{j}=27^{k}$$
$$(3^{\frac{1}{2}})^{j}=(3^{3})^{k}$$
$$\frac{j}{2}=3k$$
$$j=2\times3k$$
$$\frac{j}{k}=6$$ …
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Choice A
For an arithmetic sequence, the …
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Choice F
$$sin\theta=\frac{1}{4}$$
$$\Rightarrow sin(\frac{1}{4})^{-1}=\theta$$
$$arc\ length=2\pi …
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Choice D
The asymptote of a function …
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Choice F
Conic sections are the result …
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