Write an algebraic expression for each situation. a. 7 more than a number b. Twice a number
1 answer:
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Simple, what you're basically doing is simplifying,
well, you have 3x and you have
12/3=4
and you can "cancel" out an x, making your answer,
.
well, you have 3x and you have
12/3=4
and you can "cancel" out an x, making your answer,
<span>4. Simplify the expression.
sine of x to the second power minus one divided by cosine of negative x</span>
</span>sin2x+cos2x=1</span> <span>1−sin2x=cos2x<span>
</span>cos2(x)/(sin(x)−csc(x))</span> <span>csc(x)=1/sin(x)</span> <span>cos2(x)/(sin(x)− 1/sin(x))= cos2(x)/((sin2(x)− 1)/sin(x))</span> <span>sin2(x)− 1=-cos2(x)</span> <span>cos2(x)/(( -cos2(x))/sin(x))
=-sin(x)</span>
<span>the answer is the letter a) -sin x
</span><span>5. Find all solutions in the interval [0, 2π). (6 points)
sin2x + sin x = 0
</span> using a graphical tool
the solutions x1=0 x2=pi <span>x3=3pi/2
the answer is the letter </span><span>D) x = 0, π, three pi divided by two</span>
