# Algebra Problems To Solve

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$$240-150=90$$ Then we find out how many percent this change corresponds to when compared to the original number of students $$a=r\cdot b$$ $$90=r\cdot 150$$ $$\frac=r$$ $$0.6=r= 60\%$$ We begin by finding the ratio between the old value (the original value) and the new value $$percent\:of\:change=\frac=\frac=1.6$$ As you might remember 100% = 1.

Since we have a percent of change that is bigger than 1 we know that we have an increase.

To solve problems with percent we use the percent proportion shown in "Proportions and percent".

$$\frac=\frac$$ $$\frac\cdot =\frac\cdot b$$ $$a=\frac\cdot b$$ x/100 is called the rate.

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For example, will the answer be in miles, feet, ounces, pesos, dollars, the number of trees, or a number of televisions?

Let's try θ = 30°: sin(−30°) = −0.5 and −sin(30°) = −0.5 So it is true for θ = 30° Let's try θ = 90°: sin(−90°) = −1 and −sin(90°) = −1 So it is also true for θ = 90° Is it true for all values of θ?

How many students in the class have either glasses or contacts?

$$a=r\cdot b$$ $\%=0.47a=0.47\cdot 34a=15.98\approx 16$\$ 16 of the students wear either glasses or contacts.

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