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Sloan [31]
4 years ago
5

Please help with these two math questions quick!! (More specifically the bottom one)

Mathematics
1 answer:
Varvara68 [4.7K]4 years ago
5 0
The top one is a and the last one is d
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Help please, 17 points for this question
chubhunter [2.5K]

Answer:

24 and 15

Step-by-step explanation:

6 0
3 years ago
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I have two proof questions and I don't really know how to do them.
goldfiish [28.3K]

\angle1\cong\angle2 corresponding angles, therefore

m∠1 = m∠2 → m∠2 = 130°

\angle2\cong\angle3 vertical angles, therefore

m∠2 = m∠3 → m∠3 = 130°


If ∠1 and ∠2 are complementary, then m∠1 + m∠2 = 90°.

If ∠2 and ∠3 are complementary, then m∠2 + m∠3 = 90°

Therfore

m\angle1 + m\angle2 = m\angle2 + m\angle3\qquad|\text{subtract}\ m\angle2\ \text{ from both sides}\\\\m\angle1 = m\angle3\to\angle1\cong\angle3

5 0
3 years ago
Triangle ABC and DEF are similar
Artist 52 [7]

Answer:

e and f is 1.34

Step-by-step explanation:

5 0
3 years ago
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Which terms and 45p^4q have a GCF of 9p^3?
vredina [299]

Answer:

The answer is 18p^3r and 63p^3

Step-by-step explanation:

G.C.F of 18p^3 r and  45p^4q is = 9p^3

18p^3r = 2*3*3*p*p*p*r

45p^4q = 3*3*5*p*p*p*q

Thus the G.C.F is 3*3*p*p*p = 9p^3

G.C.F of  63p^3 and  45p^4q is = 9p^3

63p^3 = 3*3*7*p*p*p

45p^4q = 3*3*5*p*p*p*q

Thus the G.C.F is 3*3*p*p*p = 9p^3

Therefore the answer is 18p^3r and 63p^3....

7 0
4 years ago
Please help me with this :) i give 30 points .
Tatiana [17]

Answer:

2  B  4/75

3 B 1/7

Step-by-step explanation:

4 red, 3 blue, 2 yellow, 6 orange

total = 15

2.  P(red) = red/total = 4/15

replace it

so we still have 4 red, 3 blue, 2 yellow, 6 orange

total = 15

P(blue) = blue/total = 3/15 = 1/5

P(red, replace, blue) = 4/15 * 1/5 = 4/75

3 P(orange) = orange/total = 6/15 = 2/5

not replacing it

so we  have 4 red, 3 blue, 2 yellow, 5 orange

total = 14

P(orange) = orange/total = 5/14

P(orange, no replace, orange) = 2/5*5/14 = 2/14 = 1/7

6 0
3 years ago
Read 2 more answers
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