Pa pahelp po dito please

Aregular polygon is drawn in a circle so that each vertex is on the circle and is connected to the center by a radius. Each of

katrin [286]

Answer:

The number of sides the polygon have 9

Step-by-step explanation:

Given in question as :

For The circle , the central angles measure be 40 °

Let the number of sides of polygon = n

Now . as we know

External angle = $\frac{360^{\circ} }{n}$

Or , 40° = $\frac{360^{\circ} }{n}$

Or, $\frac{360^{\circ} }{n}$ = 40°

So, n = $\frac{360^{\circ} }{40^{\circ}}$ = 9

Hence The number of sides the polygon have 9 Answer

7 0

4 years ago

Complete the table of values for y=2x squared + x

Lapatulllka [165]

Answer:

y=2x^2 + 2x - 3 x -2 -1 0 1 2

Step By Step Explanation:

The values can be find by plugging the x values in the equation which gives us the y value ,

x -2 -1 0 1 2

y 1 -3 -3 1 9

(b) Plot the points on the graph and join by a smooth curve.

(c) The line y=1 will be passing through 1 and parallel to x-axis.

(d) solve 2x^2 + 2x - 3 = 1

Subtract 1 from both the sides ,

2x^2 + 2x - 2 = 0

Factoring out the 2 from the equation ,

2(x^2 + x - 1) = 0

x^2 + x - 1 = 0

Apply the quadratic formula

x=(-1-sqrt(5))/2 and x = (-1+sqrt(5))/5

5 0

3 years ago

Expression "-3x^2+y4x" when x = "-4" and y = 2 A. -112 B. -80 C. 80 D. 272

Paraphin [41]

B, -3(-4)^2+(2)4(-4) plug into a calculator

6 0

3 years ago

What is the remainder of (4x2 + 7x-1)= (4 + x)? A. -9x – 1 B.23x – 1 C.35 D.-37

bearhunter [10]

Answer: I don't know what you meant by remainder but i hope this helps :)

$x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}\\$

Step-by-step explanation:

$\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}$

$\frac{-6+\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\=\frac{-6+\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{2\times \:4}\\=\frac{-6+\sqrt{116}}{8}\\\\Let\: simplify\: ; -6+2\sqrt{29}\\=-2\times \:3+2\sqrt{29}\\=2\left(-3+\sqrt{29}\right)\\=\frac{2\left(-3+\sqrt{29}\right)}{8}\\=\frac{-3+\sqrt{29}}{4}\\$

$\frac{-6-\sqrt{6^2-4\times \:4\left(-5\right)}}{2\times \:4}\\\\=\frac{-6-\sqrt{6^2+4\times \:4\times \:5}}{2\times \:4}\\\\=\frac{-6-\sqrt{116}}{2\times \:4}\\\\=\frac{-6-2\sqrt{29}}{8}\\\\=-\frac{2\left(3+\sqrt{29}\right)}{8}\\\\=-\frac{3+\sqrt{29}}{4}\\\\\\x=\frac{-3+\sqrt{29}}{4},\:x=-\frac{3+\sqrt{29}}{4}$