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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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Mamont248 [21]

Answer:

12 gallons of yellow paint

4 gallons of blue paint

Step-by-step explanation:

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2 years ago

Is 5.35 a repeating decimal

Naya [18.7K]

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3 years ago

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Use the given information to find (a) sin(s+t), (b) tan(s+t), and (c) the quadrant of s+t. cos s = - 12/13 and sin t = 4/5, s an

Anton [14]

Answer:

Part a) $sin(s + t) =-\frac{63}{65}$

Part b) $tan(s + t) = -\frac{63}{16}$

Part c) (s+t) lie on Quadrant IV

Step-by-step explanation:

[Part a) Find sin(s+t)

we know that

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

step 1

Find sin(s)

$sin^{2}(s)+cos^{2}(s)=1$

we have

$cos(s)=-\frac{12}{13}$

substitute

$sin^{2}(s)+(-\frac{12}{13})^{2}=1$

$sin^{2}(s)+(\frac{144}{169})=1$

$sin^{2}(s)=1-(\frac{144}{169})$

$sin^{2}(s)=(\frac{25}{169})$

$sin(s)=\frac{5}{13}$ ---> is positive because s lie on II Quadrant

step 2

Find cos(t)

$sin^{2}(t)+cos^{2}(t)=1$

we have

$sin(t)=\frac{4}{5}$

substitute

$(\frac{4}{5})^{2}+cos^{2}(t)=1$

$(\frac{16}{25})+cos^{2}(t)=1$

$cos^{2}(t)=1-(\frac{16}{25})$

$cos^{2}(t)=\frac{9}{25}$

$cos(t)=-\frac{3}{5}$ is negative because t lie on II Quadrant

step 3

Find sin(s+t)

$sin(s + t) = sin(s) cos(t) + sin(t)cos(s)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute the values

$sin(s + t) = (\frac{5}{13})(-\frac{3}{5}) + (\frac{4}{5})(-\frac{12}{13})$

$sin(s + t) = -(\frac{15}{65}) -(\frac{48}{65})$

$sin(s + t) =-\frac{63}{65}$

Part b) Find tan(s+t)

we know that

tex]tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))[/tex]

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

step 1

Find tan(s)

$tan(s)=sin(s)/cos(s)$

substitute

$tan(s)=(\frac{5}{13})/(-\frac{12}{13})=-\frac{5}{12}$

step 2

Find tan(t)

$tan(t)=sin(t)/cos(t)$

substitute

$tan(t)=(\frac{4}{5})/(-\frac{3}{5})=-\frac{4}{3}$

step 3

Find tan(s+t)

$tan(s + t) = (tan(s) + tan(t))/(1 - tan(s)tan(t))$

substitute the values

$tan(s + t) = (-\frac{5}{12} -\frac{4}{3})/(1 - (-\frac{5}{12})(-\frac{4}{3}))$

$tan(s + t) = (-\frac{21}{12})/(1 - \frac{20}{36})$

$tan(s + t) = (-\frac{21}{12})/(\frac{16}{36})$

$tan(s + t) = -\frac{63}{16}$

Part c) Quadrant of s+t

we know that

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

Find the value of cos(s+t)

$cos(s+t) = cos(s) cos(t) -sin (s) sin(t)$

we have

$sin(s)=\frac{5}{13}$

$cos(t)=-\frac{3}{5}$

$sin(t)=\frac{4}{5}$

$cos(s)=-\frac{12}{13}$

substitute

$cos(s+t) = (-\frac{12}{13})(-\frac{3}{5})-(\frac{5}{13})(\frac{4}{5})$

$cos(s+t) = (\frac{36}{65})-(\frac{20}{65})$

$cos(s+t) =\frac{16}{65}$

we have that

$cos(s+t)=positive$ -----> (s+t) could be in I or IV quadrant

$sin(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

$tan(s + t) =negative$ ----> (s+t) could be in III or IV quadrant

therefore

(s+t) lie on Quadrant IV

4 0

3 years ago

Negative X +3+ 3X equals 23

Nadusha1986 [10]

Answer:

X = 10

Step-by-step explanation:

First you subtract -X from 3x to get 2x +3 =23

then you subtract 3 from both sides to get 2x = 20

then finally, you divide both sides by two to get x = 10

And if you plug 10 back in you get -10 + 3 + 30 = 23

-10 + 3 = -7

-7 + 30=23

23=23

6 0

3 years ago

What is the value of y?

vladimir1956 [14]

Answer:

65 degrees

Step-by-step explanation:

First, solve for the unknown angle( not the y's).

the angle 130 is supplementary to that angle so we will use an equation to solve it. (supplementary means it adds up to 180 degrees)

x+130=180

-130 -130

x = 50 so, 50 is the measure of the interior angle

Now we have to use another equation to solve for y:

50+2y=180

-50 -50

2y = 130

/2 /2

y = 65

Thus, y is 65

6 0

3 years ago

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