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3 years ago

Unit4: Congruent Triangles Hw 1: Classifying Triangles Gina Wilson (All things algebra) 2014

Mathematics

2 answers:

anastassius [24]3 years ago

6 0

Answer:

12. x = 6, side = 83
13. x = 18, side = 29
14. x = 11, sides = 74, 74 and 37
15. x = 23, sides = 95, 95 and 108

Step-by-step explanation:

11 is incomplete, can't solve

The triangle is equilateral, so all sides are equal, using one pair to find x:

13x + 5 = 17x - 19
17x - 13x = 5 + 19
4x = 24
x = 6

Each side is:

13*6 + 5 = 78 + 5 = 83

13.

Sides are equal as triangle is equilateral

QR = 2x - 7
RS = 5x - 61
QS = x + 11

Finding x by comparing two sides

2x - 7 = 5x - 61
5x - 2x = 61 - 7
3x = 54
x = 18

Sides are equal

18 + 11 = 29

14.

Equal sides of isosceles triangle:

CD = DE
9x - 25 = 6x + 8
9x - 6x = 8 + 25
3x = 33
x = 11

Sides are

CD = DE = 9*11 - 25 = 99 - 25 = 74
CE = 10*11 - 73 = 110 - 73 = 37

15.

Equal sides of isosceles triangle WXY, WX = WY

WX = 4x + 3
WY = 7x - 66
XY = 5x - 7
4x + 3 = 7x - 66
7x - 4x = 3 + 66
3x = 69
x = 23

Sides are:

WX=WY = 4*23 + 3 = 95
XY = 5*23 - 7 = 108

bulgar [2K]3 years ago

4 0

Answer:

11. Isosceles

Step-by-step explanation:

DE = √(-2-5)^2 +(3-5)^2 = √(-7)^2 + (-2)^2 = √49+4 = √53

DF = √(-2+4)^2 + (3-10)^2 = √(2)^2 + (-7)^2 = √4+49 = √53

EF = √(5+4)^2 + (5-10)^2 = √(9)2 + (-5)^2 = √81+25 = √106

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artcher [175]

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Step-by-step explanation:

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2 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

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2 years ago

Fill in the blank (correctly). 20 points each.

lyudmila [28]

Distributing same key numbers

7 0

3 years ago

Find the value of x if a,b, and c are collinear points an b is between a and c

MakcuM [25]

$\sf Hello!$

• $\sf AB = x + 6$
• $\sf BC = 3x - 5$
• $\sf AC = 36 - x$

$\sf We're \:given\: that \:A,\: B\: and \:C \:are \:Collinear\: points$

$\sf Then,$

$\sf AB + BC = AC$

⇒ $\sf x + 6 + 3x - 5 = 36 - x$

⇒ $\sf x + 3x + 6 - 5 = 36 - x$

⇒ $\sf 4x + 1 = 36 - x$

⇒ $\sf 4x + x = 36 - 1$

⇒ $\sf 5x = 35$

⇒ $\sf x = \dfrac{\sf 35}{\sf 5}$

⇒ $\boxed{\sf x = 7}$

$\sf Hence,$

$\sf The\: value\: of\: x\: is\: D.\: 7$

~ $\sf iCarl$

3 0

3 years ago

A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local b

IrinaK [193]

Answer:

a. [ 0.454,0.51]

b. 599.472 ~ 600

Step-by-step explanation:

Confidence Interval For Proportion

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))

x = Mean

n = Sample Size

a = 1 - (Confidence Level/100)

Za/2 = Z-table value

CI = Confidence Interval

Mean(x)=410

Sample Size(n)=850

Sample proportion = x/n =0.482

Confidence Interval = [ 0.482 ±Z a/2 ( Sqrt ( 0.482*0.518) /850)]

= [ 0.482 - 1.645* Sqrt(0) , 0.482 + 1.65* Sqrt(0) ]

= [ 0.454,0.51]

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.05 is = 1.96

Samle Proportion = 0.482

ME = 0.04

n = ( 1.96 / 0.04 )^2 * 0.482*0.518

= 599.472 ~ 600

8 0

3 years ago