Simplify the following √5 x 2√2

Mathematics

1 answer:

seraphim [82]2 years ago

5 0

Answer:

try mopo quanda or guthmath

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More on Areas. Farmer Jones, and his wife, Dr. Jones, both mathematicians, decide to build a fence in their field to keep the sh

Olin [163]

Answer:

Area covered by the fences will be 16.1 unit²

Step-by-step explanation:

Let the first parabola is represented by the function f(x) = 6x²

and second parabola by g(x) = x² + 9

point of intersection of the graphs will be determined when f(x) = g(x)

6x² = x² + 9

5x² = 9

x² = 1.8

x = ± 1.34

Now we will find the area between these curves drawn on the graph.

Area = $\int_{-1.34}^{1.34}[f(x)-g(x)]dx=\int_{-1.34}^{1.34}[6x^{2}-(x^{2}+9)]dx$

= $\int_{-1.34}^{1.34}(5x^{2}-9)dx$

= $[\frac{5}{3}x^{3}-9x]_{-1.34}^{1.34}$

= $[\frac{5}{3}(-1.34)^{3}-9(-1.34)-\frac{5}{3}(1.34)^{3}+9(1.34)]$

= $[-4.01+12.06-4.01+12.06]$

= 16.1 unit²

6 0

3 years ago

The function f(x) is the total amount spent at a store, when purchasing x items that are $5 each and the items are not taxable

Keith_Richards [23]

F(x)=5x

normal domain: all real numbers
practical domain: <span>all positive integers

</span>becasue we can substituent with any positive integer in the place of x

3 0

3 years ago

Read 2 more answers

Help with Pre Calculus

marshall27 [118]

Answer:

The first option

Step-by-step explanation:

The domain of a rational function should be all real numbers except for when the denominator is equal to 0. To find when the denominator is equal to 0 you simply need to find the zeroes of the denominator... but in this case you can do that through factoring and using the quadratic equation.

So first step is going to be to factor out the GCF, which in this case is x. This gives you the equation. $x(2x^2-x-15)$ . So one of the zeroes is when x=0. Now to find the other two zeroes you can use the quadratic equation which is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\$ . So to find the other zeroes you simply plug the values in. a=2, b=-1, c=-15

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(2)(-15)}}{2(2)}\\\\x=\frac{1\pm\sqrt{1-4(2)(-15)}}{4}\\\\x=\frac{1\pm\sqrt{121)}}{4}\\\\x=\frac{1\pm11}{4}\\\\x=\frac{12}{4}\\x=3\\\\x=\frac{-10}{4}\\x=-\frac{5}{2}$