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Y_Kistochka [10]

4 years ago

9

Use graph paper and a straightedge to draw the figure. Which best describes the polygon whose vertices in the coordinate plane a

re (–5, 5), (0, 5), (0, 0), (–5, 0)? A.
only equilateral

B.
only equiangular

C.
regular
PLEASE HURRY!

1 answer:

Rama09 [41]4 years ago

3 0

C i think but im not sure

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jok3333 [9.3K]

Answer:

1.-4,-1,1,6

2.by substituting the values given, we get

32-|15+8|

32-|23|

32-23=9

3.-15+7=8

4..27-(-12)=27+12=39

5.-6.05+(-2.1)

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-3/4+2/5

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8.(3.8)(-4.1)

-15.58

9.(-8x)(-2y)+(-3y)(z)

(16xy)+(-3zy)

(16xy)-(3zy)

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7 0

3 years ago

Determine if the relationship is a function & explain why and why not.

SCORPION-xisa [38]

It is not a function because there can't be two of the same number on the x-axis.

7 0

4 years ago

What’s the differential equation of

worty [1.4K]

Answer:

$y=\frac{1}{x^{2}-6x+13 }$

Step-by-step explanation:

We have given,

$\frac{dy}{dx}=y^{2}(6-2x)$

and initial condition $x=3,\ y=\frac{1}{4}$

Now,

$\frac{dy}{dx}=y^{2}(6-2x)$

Rearranging the variables, we get

$\frac{dy}{y^{2} }=(6-2x).dx$

Applying integration both sides, we get

$\int\ {\frac{dy}{y^{2} } } \,=\int\ {(6-2x).} \, dx$

⇒ $\frac{-1}{y} =6x-\frac{2x^{2} }{2}$

⇒ $\frac{-1}{y}=6x-x^{2} +C$

Putting the initial condition (i.e., $x=3,\ y=\frac{1}{4}$ ), we get

⇒ $-4=6\times3-(3)^{2}+C$

⇒ $-4=18-9+C$

∴ $C=-13$

We have, $\frac{-1}{y}=6x-x^{2} +C$

now putting the value of $C$ in above equation, we get

⇒ $\frac{-1}{y}=6x-x^{2} -13$

⇒ $\frac{1}{y}=-6x+x^{2} +13$

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3 years ago

Find the measure of the missing angles.<br> b<br> 100°<br> 0

Jobisdone [24]

Step-by-step explanation:

Angle b + 100 = 180 (angles on the same straight line)

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Angle c = Angle b (Vertically opposite angles)

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3 years ago

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Juli2301 [7.4K]

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