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2 years ago

15

I have more sides than a quadrilateral , but fewer sides than a hexagon who could I be

2 answers:

Korolek [52]2 years ago

7 0

You would be a pentagon. A pentagon has 5 sides

kirill115 [55]2 years ago

3 0

A pentagon. A pentagon has 5 sides.

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NEED HELP ASAP WILL MARK BRAINLET FOR GOOD AND CORRECT ANSWER!!!!

hjlf

Hey! The surface area is 294. Have a great day. :-)

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2 years ago

ILL GIVE BRAINLIST! plus 12 points!

kozerog [31]

Answer:

B

Step-by-step explanation:

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2 years ago

Read 2 more answers

Explain in detail how to find the vertex: <br> y=-2x^2 + 2x + 3

Cerrena [4.2K]

$\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-2}x^2\stackrel{\stackrel{b}{\downarrow }}{+2}x\stackrel{\stackrel{c}{\downarrow }}{+3} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\bf \left(-\cfrac{2}{2(-2)}~~,~~3-\cfrac{2^2}{4(-2)} \right)\implies \left( \cfrac{1}{2}~~,~~3+\cfrac{4}{8} \right)\implies \left(\cfrac{1}{2}~~,~~\cfrac{28}{8} \right) \\\\\\ \left(\cfrac{1}{2}~~,~~\cfrac{7}{2} \right)\implies \left(\frac{1}{2}~,~3\frac{1}{2} \right)$

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2 years ago

What are the exact solutions of x2 + 5x + 1 = 0? (5 points)

Rus_ich [418]

Answer:

$x=\frac{-5\pm\sqrt{21} }{2}$

Step-by-step explanation:

We use the quadratic formula here which says for a quadratic equation $ax^2+bx+c=0$ .

$x= \frac{-b\pm\sqrt{b^2-4ac} }{2a}$

Now in our case

$a=1\\b=5\\c=1$

so we have:

$x=\frac{-5\pm\sqrt{5^2-4(1)(1)} }{2(1)} =\frac{-5\pm\sqrt{21} }{2}$

$\boxed{x=\frac{-5\pm\sqrt{21} }{2} }$

Which are our solutions.

7 0

3 years ago

5, -10, 20, ...; 6th term<br> What is the 6th term?

aksik [14]

Answer:

20

Step-by-step explanation:

To get to -10 from 5 you have to subtract 15

to get to 20 from -10 you have to add 30

Repeat that process until you get to the 6th term.

5 (1st term)

5 - 15 = -10 (2nd term)

-10 + 30 = 20 (3rd term)

20 - 15 = 5 (4th term)

5 + 30 = 35 (5th term)

35 - 15 = 20 (6th term)

5 0

2 years ago