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Sedaia [141]
3 years ago
13

the diagonals of parallelogram ABCD intersect at point O. If segment AC is a diagonal and AO = 5, then what is AC

Mathematics
1 answer:
gavmur [86]3 years ago
8 0

9514 1404 393

Answer:

  10

Step-by-step explanation:

The diagonals of a parallelogram bisect each other. That is, point O is the midpoint of AC. If AO = 5, then OC = 5 and AC = 5+5 = 10.

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Miguel napped for 2 2/5 hours. Danalya napped too. Combined they napped for 3 3/4 hours
SVEN [57.7K]

Answer:

Danalya napped 1 7/20 hours.

Step-by-step explanation:

Rewriting our equation with parts separated

=3+3/4−2−2/5

Solving the whole number parts

3−2=1

Solving the fraction parts

3/4−2/5=?

Find the LCD of 3/4 and 2/5 and rewrite to solve with the equivalent fractions.

LCD = 20

15/20−8/20= 7/20

Combining the whole and fraction parts

1+7/20= 1 7/20

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3 years ago
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Solve the compound inequality. Graph the solution<br> -2 ≤ 2x – 4 &lt; 4
Ulleksa [173]
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&#10;1\ \textless \ x\ \textless \ 4&#10; &#10; = (1,4)
5 0
3 years ago
What is the solution to the inequality shown? -2(x + 3) &lt; 6 - x x &gt; -12 x &lt; -12 x &gt; 12 x &lt; 12
Sonja [21]

Answer:

x < -12

Step-by-step explanation:

Given the inequality

-2(x + 3) < 6 - x

Expand

-2x - 6 < 6 - x

Collect the like terms;

-2x + x < 6 + 6

-x < 12

Multiply through by -1

-1(-x) < -1(12)

x < -12

6 0
3 years ago
A museum requires a minimum number of chaperones proportional to the number of students on a field trip. The museum requires a m
Stells [14]

Answer: See explanation

Step-by-step explanation:

Here is the complete question:

A museum requires a minimum number of chaperones proportional to the number of students on a field trip. The museum requires a minimum of 3 chaperones for a field trip with 24 students. Which of the following could be combinations of values for the students and the minimum number of chaperones the museum requires? Choose 2 answers.

A. Students: 72

Minimum of chaperones: 9

B. Students: 16

Minimum of chaperones: 2

C. Students: 60

Minimum of chaperones: 6

D. Students: 45

Minimum of chaperones: 5

E. Students: 40

Minimum of chaperones: 8

Since the museum requires a minimum of 3 chaperones for a field trip with 24 students. This means that there will be 24/3 = 8 students per chaperone.

We then divide the number of students given in the question by the number of chaperone to know our answers. This. Will be:

Students: 72

Minimum of chaperones: 9

This will be: 72/9 = 8

Therefore, this is correct.

B. Students: 16

Minimum of chaperones: 2

This will be: 16/2 = 8

This is correct

C. Students: 60

Minimum of chaperones: 6

This will be: = 60/6 = 10.

Therefore, this is wrong

D. Students: 45

Minimum of chaperones: 5

This will be 45/5 = 9

Therefore, this is wrong.

E. Students: 40

Minimum of chaperones: 8

This will be: 40/5 = 8.

Therefore, this is wrong.

Therefore, options A and B are correct.

3 0
3 years ago
Read 2 more answers
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andrew11 [14]

Answer:

21 per min

Step-by-step explanation:

8 0
3 years ago
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