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tekilochka [14]
3 years ago
8

What is the equation of a line passing through the points 6,3 and -4,3 so y=?

Mathematics
1 answer:
Mice21 [21]3 years ago
7 0
Equation of a line using 2 points. first find the slope

m= (y2-y1)/(x2-x1)= (3-3)/(-4-6)= 0/-10 = 0
m=0

using y-y1=m(x-x1)
we get
y-3=0(x-6)
y-3=0

y=3 is the equation of the line
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Given: ΔWXY is isosceles with legs WX and WY; ΔWVZ is isosceles with legs WV and WZ. Prove: ΔWXY ~ ΔWVZ
Alex777 [14]

Answer:

By AA

ΔWXY ~ΔWVZ

Step-by-step explanation:

Here WXY is an isosceles triangle with legs WX & WY

So WX = WY

Hence ∠X = ∠Y

So ∠2= ∠3.

Now by angle sum property

∠1 + ∠2+∠3 = 180°

∠1+∠2+∠2=180°

2∠2 = 180° - ∠1     .......(1)

In triangle WVZ

WV = WZ

So ∠V = ∠Z

∠4 = ∠5

Once again by angle sum property

∠1 + ∠4 + ∠5=180°

∠1 + ∠4 + ∠4 = 180°

2∠4 = 180° - ∠1       ...(2)

From (1) & (2)

2∠2 = 2∠4

∠2=∠4

Now ∠W is common to both triangles

Hence by AA

ΔWXY ~ΔWVZ

8 0
3 years ago
Read 2 more answers
A square garden has a side length of 11 meters. What is the area of the garden? What is the perimeter?
Talja [164]

Answer:

<em>Area</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>garden</em><em>=</em><em>1</em><em>2</em><em>1</em><em> </em><em>square</em><em> </em><em>meter</em>

<em>perimeter</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>garden</em><em>=</em><em>4</em><em>4</em><em> </em><em>meter</em>

Step-by-step explanation:

Area of a square=side*side

=11*11

=<u>1</u><u>2</u><u>1</u><em><u>s</u></em><em><u>q</u></em><em><u>u</u></em><em><u>a</u></em><em><u>r</u></em><em><u>e</u></em><em><u> </u></em><em><u>meter</u></em>

Perimeter of a square garden=side*4

=11*4

=44 m

6 0
3 years ago
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Colton has n nickels and d dimes. He has no more
pantera1 [17]

Answer:

(`n*0.05)+(d*0.10)<2

Step-by-step explanation:

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3 years ago
Find two rational expressions a / b and c / d that produce the result x − 1 / x2 when using the following operations. Answers
Mars2501 [29]

Answer:

a) Let \frac{a}{b}=\frac{-1}{x^2}, \text{ and } \frac{c}{d}=\frac{1}{x}.

Observe that

\frac{a}{b}+\frac{c}{d}=\frac{-1}{x^2}+\frac{1}{x}=\frac{-x+x^2}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

b)

Let \frac{a}{b}=\frac{1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x^2}.

Observe that

\frac{a}{b}-\frac{c}{d}=\frac{1}{x}-\frac{1}{x^2}=\frac{x^2-x}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}

c)

Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x}.

Observe that

\frac{a}{b}*\frac{c}{d}=\frac{x-1}{x}*\frac{1}{x}=\frac{(x-1)1}{x*x}=\frac{x-1}{x^2}

d)

Let \frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{x}{1}.

Observe that

\frac{a}{b}\div\frac{c}{d}=\frac{x-1}{x}\div\frac{x}{1}=\frac{x-1}{x}*\frac{1}{x}=\frac{x-1}{x^2}

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