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LenKa [72]
3 years ago
10

Are the triangles similar? How do you know? A. yes, by SAS B. yes, by SSS C. yes, by AA D. no

Mathematics
2 answers:
lord [1]3 years ago
8 0

Answer:

Option D is correct.

No, the triangles are not similar.

Step-by-step explanation:

AA postulates states that two triangles are similar if they have two corresponding angles equal.

Labelled the diagram as shown below:

We know that sum of all the measure of the angles in a triangle is 180 degree.

In triangle  ABC:

\angle A + \angle B + \angle C = 180^{\circ}

⇒30.4^{\circ}+84.6^{\circ}+\angle C = 180^{\circ}

⇒115^{\circ}+\angle C = 180^{\circ}

Subtract 115 degree from both sides we get;

⇒\angle C = 65^{\circ}

In triangle ABC and PQR

\angle A = \angle Q = 84.6^{\circ}

\angle C \neq \angle R

i.e 65^{\circ}\neq 66^{\circ}

⇒These triangles does not satisfy the AA postulates

Therefore, the given triangles are not similar.

WARRIOR [948]3 years ago
4 0
No, because 30.4+84.6+66=181
the sum of a triangle=180
so they are not similar. 
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sukhopar [10]
She buys 2 boxes each week for a year (there is 52 weeks in a year) = 2 x 52 = 104 boxes per year
if each box contains 8 granola bars...and she buys 104 boxes, then she buys:
104 * 8 = 832 granola bars
6 0
3 years ago
Can I get help solving this graph please?
Daniel [21]
see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100)  B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40)  C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0)  B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60)  C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then 
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x) 
</span>h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then 
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x) 
h'(x)=18/25=0.72 

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

8 0
3 years ago
A salesperson at a jewelry store earns 4​% commission each week. Last​ week, Heidi sold $680 worth of jewelry. How much did make
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Answer:

$27.2

Step-by-step explanation:

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48 / 8 = 6.

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