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3 years ago

2. A polygon that has no interior angles greater than 180 degrees is a ________________ polygon. *

Mathematics

2 answers:

azamat3 years ago

4 0

Answer: convex

Step-by-step explanation:

dalvyx [7]3 years ago

4 0

Your answer = convex

hope this helps:)

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Once simplified, which of the expressions below have a value greater than -20?

balandron [24]

Answer:

A, B, and C

Step-by-step explanation:

a). -6 ÷ 2 x 5 1/3

= -3 x 5 1/3

= -16

b). 24 ÷ (-3) + 1/2

= -8 + 1/2

= -7.5

c). 36 ÷ 4 x (-2)

= 9 x -2

= -18

d). -54 ÷ (-9) x (-3 2/3)

6 x -3 2/3

= -22

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2 years ago

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Dominik [7]

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4 years ago

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Ede4ka [16]

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3 years ago

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Verify identity: <br><br> (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)

Nikitich [7]

So hmmm let's do the left-hand-side first

$\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies 
\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

now, let's do the right-hand-side then

$\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}$

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3 years ago

What is the value of f if f + I-3 11/12 = 18

astra-53 [7]

Solve for f:(i - 47/12) + f = 18
(i - 47/12) + f = f + (-47/12 + i):
f + (-47/12 + i) = 18
Subtract -47/12 + i from both sides:
Answer: f = 263/12 - i

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3 years ago