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natali 33 [55]
2 years ago
11

Is this shape a regular polygon

Mathematics
2 answers:
alex41 [277]2 years ago
8 0
No that’s an octagon
Anettt [7]2 years ago
6 0
It is not a polygon, it is a octagon. Oct- is 8 and that shape has 8 sides
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Does anyone know the answers to this test???OFFERING LOTS PF POINTS. Just Incase the picture isn’t loading it’s the parametric f
steposvetlana [31]

Answer:

The correct choice is C

Step-by-step explanation:

The given curve is described by the parametric equations:

x=4-t

y=t^2-2

Let us eliminate the parameter by making t the subject in the first equation and substitute into the second equation;

t=4-x

We substitute this into the second equation to get:

y=(4-x)^2-2

This is the equation of a parabola whose vertex is at (4,-2)

The correct choice is C

6 0
3 years ago
What is 40% of 42?<br> Please answer fast
Hunter-Best [27]

Answer:

16.8 is the answer

Step-by-step explanation:

(.40 x 42= 16.8)

It's super simple. Hope this helped!

8 0
3 years ago
Read 2 more answers
42/35 in simpliest form
OverLord2011 [107]
The simpliest form is 6/5
4 0
3 years ago
Find the values of x and y for the isosceles trapezoid
butalik [34]
<h3>Answer:</h3>
  • x = 15
  • y = 3
<h3>Step-by-step explanation:</h3>

If the trapezoid is isosceles, angles A and C are supplementary, so ...

... (4x+4) +(7x+11) = 180

... 11x +15 = 180 . . . . . . collect terms

... 11x = 165 . . . . . . . . . subtract 15

... x = 15 . . . . . . . . . . . . divide by the coefficient of x

And angles C and E are congruent.

... 4·15 +4 = 21y +1

... 63 = 21y . . . . . subtract 1

... 3 = y . . . . . . . . divide by the coefficient of y

4 0
3 years ago
Solve 2x2 + 20x = −38. (1 point)
RoseWind [281]

Answer:

The solutions are x=-5+\sqrt{6}  and x=-5-\sqrt{6}


Step-by-step explanation:

we have

2x^{2} +20x=-38

Divide by 2 both sides

x^{2} +10x=-19 ------> x^{2} +10x+19=0

we know that


The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to


x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}


in this problem we have


x^{2} +10x+19=0

so


a=1\\b=10\\c=19


substitute

x=\frac{-10(+/-)\sqrt{10^{2}-4(1)(19)}}{2(1)}


x=\frac{-10(+/-)\sqrt{100-76}}{2}


x=\frac{-10(+/-)\sqrt{24}}{2}


x=\frac{-10(+/-)2\sqrt{6}}{2}


x1=\frac{-10(+)2\sqrt{6}}{2}=-5+\sqrt{6}


x2=\frac{-10(-)2\sqrt{6}}{2}=-5-\sqrt{6}


8 0
3 years ago
Read 2 more answers
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