Chapter 5: Problem 91
Evaluate. $$ 38 \div 10 $$
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Chapter 5: Problem 91
Evaluate. $$ 38 \div 10 $$
These are the key concepts you need to understand to accurately answer the question.
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Evaluate. $$ 1000(1.53) $$
Multiply. $$ \frac{1}{2}(4 m-8 n) $$
The special product $$ (x+y)(x-y)=x^{2}-y^{2} $$ $$ \text { can be used to perform some multiplication problems. Here are two examples.} $$ $$ \begin{aligned} 51 \times 49 &=(50+1)(50-1) \\ &=50^{2}-1^{2} \\ &=2500-1^{2} \\ &=2499 \end{aligned} \quad | \begin{aligned} 102 \times 98 &=(100+2)(100-2) \\ &=100^{2}-2^{2} \\ &=10,000-4 \\ &=9996 \end{aligned} $$ Once these patterns are recognized, multiplications of this type can be done mentally. Use this method to calculate each product mentally. $$ 301 \times 299 $$
The polynomial equation $$ y=-0.0545 x^{2}+5.047 x+11.78 $$ gives a good approximation of the age of a dog in human years y, where \(x\) represents age in dog years. Each time we evaluate this polynomial for a value of \(x,\) we get one and only one output value y. For example, if a dogs is 4 in dog years, let \(x=4\) to find that \(y=31.1\) (lirify this, This means that the dogs is about 31 yr old in human years. This illustrates the concept of a function, one of the most important topics in mathematics. It used to be thought that each dog year was about 7 human years, so that \(y=7 x\) gave the number of human years for \(x\) dog years. Evaluate \(y\) for \(x=9,\) and interpret the result.
If an object is projected upward under certain conditions, its height in feet is given by the trinomial $$ -16 x^{2}+60 x+80 $$ where \(x\) is in seconds. Evaluate this polynomial for \(x=2.5 .\) Use the result to fill in the blanks: If___ \(-\) seconds have elapsed, the height of the object is __ feet.
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