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Tasya [4]
3 years ago
9

What is 36÷-12+5 (20-5)

Mathematics
2 answers:
Furkat [3]3 years ago
7 0
<span>36÷-12+5 (20-5)= </span>72 is the answer to your problem
Bas_tet [7]3 years ago
4 0
36 ÷ (-12) +5(20-5)
=36÷(-12) +5(15)
=36÷(-12)+75
=-3+75
=72
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The sequence is -6,-2,2,6,10 find a20​
Keith_Richards [23]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Yes, it is correct
alexgriva [62]

Answer:In Heather's solution to the problem, she wrote and solved an equation.

Her work is:

Step 1: 1.08(x +9.01 +0.98 +5.01) = 21.87

Solving like terms

1.08(x+15}=21.87

dividing both side by 1.08

we get

x+15=21.87/1.08

x+15=20.25

subtracting both side by 13.9198

x=20.25-15

x=5.25

According to correction

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<u>Option 1st</u>

<u>yes,it is correct</u>

7 0
3 years ago
URGENT please help!!! I need a explanation please!! You own a restaurant and can reopen as long as people seated at a table are
hram777 [196]

Answer:

1) Maximum number of tables that can be kept in the dinning room is 10 tables

2)The maximum number of customers, given 6 people per table is 60 customers

Step-by-step explanation:

The parameters given are;

Distance between people seated at different tables = 6 ft

Dimension of table = 2 m by 1.5 m

Number of people at a table = 6 people

Dimension of dinning;

Width = 12 m

Length = 20 m

Dimension of entrance;

Width = 20/21*30  m

Length = 10 m

With 6 meters between tables and a table with of 1.5, we have for the arrangement in the question;

3 tables = 4.5 m

Distance between = 2 × 6 = 12

4.5 + 12 = 16 m (More room required)

With two tables, we have;

Width of tables = 2×1.5 = 3 m

Distance between = 6 m

Dining room width required = 3 + 6 = 9 m

Therefore, the maximum tables in each row = 2

Given that the dining room area extends to the entrance area, total length = 30 m.

With 4 tables we have;

Length = 4 × 2 + 6 × 3 = 26

While on the entrance side of the dinning room area which is 20 m, we have 3 tables;

Length = 3 × 2 + 6 × 2 = 18

Therefore, both arrangements are acceptable;

Just on entering by the right the length = 20/21×30 m

Therefore, with 3 tables will required number due to proximity wit the door and tables arranged along the right wall of the dinning

1) Maximum number of tables that can be kept in the dinning room = 4 + 3 + 3 = 10 tables

2)The maximum number of customers, given 6 people per table = 6×10 = 60 customers.

4 0
3 years ago
Which congruence criteria can be used directly from the information about triangles HIJ and MNO to prove IJ¯¯¯¯¯¯≅MN¯¯¯¯¯¯¯¯¯¯ b
skelet666 [1.2K]

Answer: \overline{IJ}\cong \overline{MN}

Step-by-step explanation:

Given: In triangle HIJ and triangle MNO we have

\overline{HI}\cong \overline{NO}, \angle{I}\cong \angle{N} ,\angle{H}\cong \angle{O}

here, HI and NN are the included side between  ∠I & ∠H and ∠N and ∠O.

So , by ASA congruence rule,

ΔHIJ ≅ ΔMNO

So by CPCTC  (corresponding parts of the congruent triangles are congruent)

\overline{IJ}\cong \overline{MN}

7 0
3 years ago
Simplify the expression by combining like terms . 2(3t + 4) + 6(t + 2)
Talja [164]

Distribute 2 to 3t and 4, and 6 to t and 2

2(3t + 4) = 6t + 8

6(t + 2) = 6t + 12

Combine like terms:

6t + 8 + 6t + 12 = 12t + 20

12t + 20 is your answer

hope this helps

3 0
3 years ago
Read 2 more answers
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