$Solution,\\Area~of~triangle=\frac{1}{2} (base)(height) = \frac{1}{2}(2x+\frac{1}{4}) (x-8)\\~~~~~~~~~~~~~~~~~~~~~~~~ =\frac{1}{2}[2x^2-16x+\frac{1}{4}x-\frac{1}{4}(8)]\\ ~~~~~~~~~~~~~~~~~~~~~~~~ =\frac{1}{2}[2x^2-\frac{63}{4}x - 2]\\~~~~~~~~~~~~~~~~~~~~~~~~=x^2-\frac{63}{8}x-1$

7 0

3 years ago

Arrange the functions in ascending order, starting with the function that eventually has the least value and ending with the fun

natali 33 [55]

The main factor when x values are high is the nature of the function. For example, polynomial functions intrinsically grow slower than exponential functions when x is high. Also, the greater the degree of the polynomial, the more the function grows in absolute value as x goes to very large values.

In specific, this means that our 2 exponential functions grow faster than all the other functions (which are polynomial) and thus they take up the last seats. Also, 7^x grows slower than 8^x because the base is lower. Hence, the last is 8^x+3, the second to last is 7^x.

Now, we have that a polynomial of 2nd degree curves upwards faster than a linear polynomial when x is large. Hence, we have that the two 2nd degree polynomials will be growing faster than the 2 linear ones and hence we get that they fill in the middle boxes. Because x^2+4>x^2, we have that x^2+4 is the 4th from the top and x^2 is the 3rd from the top.

Finally, we need to check which of the remaining functions is larger. Now, 5x+3 is larger than 5x, so it goes to the 2nd box. Now we are done.

6 0

3 years ago