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3 years ago

Which of the following polygons has five lines of symmetry?

Mathematics

2 answers:

Natali [406]3 years ago

7 0

PENTAGON!! PENTA=5 HEPTA=7 TRI=3

Ronch [10]3 years ago

3 0

Answer:

pentagon have 5 sides, 5 symmetry lines.

So your answer is Option C.

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Which congruence theorem can be used to prove △WXS ≅ △YZS? Triangles W X S and Y Z S are connected at point S. Angles W X S and

iVinArrow [24]

Point S makes the two connected angles the same. They both have right angles.

And one side of each is congruent.

This means you know 2 angles are the same and one side is the same.

You would use ASA (Angle, Side, Angle)

3 0

3 years ago

Read 2 more answers

What is the unit of time (t) when calculating interest?

Ymorist [56]

Answer:

t = time in decimal years; e.g., 6 months is calculated as 0.5 years. Divide your partial year number of months by 12 to get the decimal years.

Step-by-step explanation:

7 0

2 years ago

Steve and Conner together have $60. Steve has $9 more than twice Conner’s amount. How much money does each have?

Lapatulllka [165]

Answer:

Step-by-step explanation:

let a=Ann's amt., b=Betty's amt.

a=2b+9

b+2b+9=60

3b+9-9=60-9

3b=51

b=$17

a=34+9=$43

6 0

3 years ago

The probability for event A is 0.3, the probability for event B is 0.6, and the probability of events A or B is 0.8. Why are the

Sedaia [141]

Answer:

Step-by-step explanation:

I think it makes the most sense sorry if i got it wrong

6 0

3 years ago

PLEASE HELP!!! Find the equation , in the standard form of the line passing through the points (3,-4) and (5,1)

ExtremeBDS [4]

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 3 &,& -4~) 
% (c,d)
&&(~ 5 &,& 1~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-4)}{5-3}\implies \cfrac{1+4}{5-3}\implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-4)=\cfrac{5}{2}(x-3)\implies y+4=\cfrac{5}{2}x-\cfrac{15}{2}$

$\bf y=\cfrac{5}{2}x-\cfrac{15}{2}-4\implies y=\cfrac{5}{2}x-\cfrac{23}{2}\impliedby 
\begin{array}{llll}
\textit{now let's multiply both}\\
\textit{sides by }\stackrel{LCD}{2}
\end{array}
\\\\\\
2(y)=2\left( \cfrac{5}{2}x-\cfrac{23}{2} \right)\implies 2y=5x-23\implies \stackrel{standard~form}{-5x+2y=-23}
\\\\\\
\textit{and if we multiply both sides by -1}\qquad 5x-2y=23$

side note: multiplying by the LCD of both sides is just to get rid of the denominators

5 0

3 years ago