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seropon [69]
2 years ago
6

4) There are 4,200 beads in a box. There are 6 bags of beads in each box. If each bag has the same number of beads, how many bea

ds are in each bag? * O 700 O 600 O 820 O 580​
Mathematics
1 answer:
Helga [31]2 years ago
4 0

Answer:

700 beads are in each bag.

Step-by-step explanation:

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I used a calculator and have solved it without. I didn't get any of these answers. HELP
Andreas93 [3]

Answer:

-9 + 7√5.

Step-by-step explanation:

(3 - √5)(2 + 3√5)

= 3(2 + 3√5) - √5(2 + 3√5))

= 6 + 9√5 - 2√5 - 15

= -9 + 7√5.

6 0
3 years ago
In a race, Andrew ran 4 minutes less than 4 times as many minutes as Chris ran. Write an expression to represent the number of m
Aleksandr [31]

Answer:

4y - 4

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Unit 1 geometry basics quiz 1-3 angle measures and relationships
ASHA 777 [7]

Answer/Step-by-step explanation:

1. Side CD and side DG meet at endpoint D to form <4. Therefore, the sides of <4 are:

Side CD and side DG.

2. Vertex of <2 is the endpoint at which two sides meet to form <2.

Vertex of <2 is D.

3. Another name for <3 is <EDG

4. <5 is less than 90°. Therefore, <5 can be classified as an acute angle.

5. <CDE is less than 180° but greater than 90°. Therefore, <CDE is classified as an obtuse angle.

6. m<5 = 42°

m<1 = 117°

m<CDF = ?

m<5 + m<1 = m<CDF (angle addition postulate)

42° + 117° = m<CDF (Substitution)

159° = m<CDF

m<CDF = 159°

7. m<3 = 73°

m<FDE = ?

m<FDG = right angle = 90°

m<3 + m<FDE = m<FDG (Angle addition postulate)

73° + m<FDE = 90° (Substitution)

73° + m<FDE - 73° = 90° - 73°

m<FDE = 17°

3 0
3 years ago
Tamu ran 30 minutes and burned 170 calories. PART A: Which equations can be used to find the average number of calories, c, burn
Lena [83]

Answer:

170 divided by 30 or calories divided by minutes

Step-by-step explanation:

the answer could be either but they both mean the same thing just in different words

8 0
2 years ago
Please Solve The Triangle for x
telo118 [61]
What triangle? Is there one on here?
4 0
2 years ago
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