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1 year ago

How do I solve “1+2+3+4+5+…100=“ A. 1010 B. 5050 C. 5000 D. 1000

Mathematics

1 answer:

irina [24]1 year ago

5 0

Answer:

5050

Step-by-step explanation:

we know that 100/2 is 50 and 50 x 100 is 5000

so now another 50 is remaining but we can't multiply but add so

1+2+3+4+5...100 = in simple form= 50x100+50=5050//

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Consider the polynomials p(x) = 3x + 27x^2 and q(x)= 2 . Find the x -coordinate(s) of the point(s) of intersection of these two

Tom [10]

Answer:

The x -coordinate(s) of the point(s) of intersection of these two polynomials are $x=\frac{2}{9}\approx0.2222,\:x=-\frac{1}{3}\approx-0.3333$

The sum of these x -coordinates is $\frac{2}{9}+\left(-\frac{1}{3}\right)=-\frac{1}{9}$

Step-by-step explanation:

The intersections of the two polynomials, p(x) and q(x), are the roots of the equation p(x) = q(x).

Thus, $3x + 27x^2=2$ and we solve for x

$3x+27x^2-2=2-2\\27x^2+3x-2=0\\\left(27x^2-6x\right)+\left(9x-2\right)\\3x\left(9x-2\right)+\left(9x-2\right)\\\left(9x-2\right)\left(3x+1\right)=0$

Using Zero Factor Theorem: = 0 if and only if = 0 or = 0

$9x-2=0\\9x=2\\x=\frac{2}{9}$

$3x+1=0\\3x=-1\\x=-\frac{1}{3}$

The solutions are:

$x=\frac{2}{9}\approx0.2222,\:x=-\frac{1}{3}\approx-0.3333$

The sum of these x -coordinates is

$\frac{2}{9}+\left(-\frac{1}{3}\right)=-\frac{1}{9}$

We can check our work with the graph of the two polynomials.

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Answer:

$2489ft^{2}$

Step-by-step explanation:

The pool are is divided into 4 separated shapes: 2 circular sections and 2 isosceles triangles. Basically, to calculate the whole area, we need to find the area of each section. Due to its symmetry, both triangles are equal, and both circular sections are also the same, so it would be enough to calculate 1 circular section and 1 triangle, then multiply it by 2.

<h3>Area of each triangle:</h3>

From the figure, we know that <em>b = 20ft </em>and <em>h = 25ft. </em>So, the area would be:

$A_{t}=\frac{b.h}{2}=\frac{(20ft)(25ft)}{2}=250ft^{2}$

<h3>Area of each circular section:</h3>

From the figure, we know that $\alpha =2.21 radians$ and the radius is $R=30ft$ . So, the are would be calculated with this formula:

$A_{cs}=\frac{\pi R^{2}\alpha}{360\°}$

Replacing all values:

$A_{cs}=\frac{(3.14)(30ft)^{2}(2.21radians)}{6.28radians}$

Remember that $360\°=6.28radians$

Therefore, $A_{cs}=994.5ft^{2}$

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