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IrinaVladis [17]
2 years ago
10

Name the property the equation illustrates. 6(-6/2) = (-6/2)6

Mathematics
2 answers:
pentagon [3]2 years ago
6 0

Answer:


Step-by-step explanation:

If you mean by solving this then okay ,

ANSWERE IS 18

blondinia [14]2 years ago
4 0

Inverse property of multiplication.

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Recently, Caroline purchased a BBQ grill to use this summer. The cost of the grill was $390 and sales tax
Dimas [21]

hi

Sales  taxes  = 390 *0.074 = 28.86  

Total cost :  390+28.86 =  418.86

7 0
2 years ago
Determine the solution set of (x - 4)2 = 12.
galben [10]
(x-4)2=12
First, you want to divide both sides by 2 to eliminate the 2 outside of the parenthesis. That leaves you with:
(x-4)=6
Then add four to both sides to eliminate the four inside the parenthesis:
x=10

To check work, substitute ten in for x in the original equation:
(10-4)2=12
(6)2=12
12=12, so this is correct :)
3 0
3 years ago
Imani was assigned to make 64 cookies for the bake sale. If Imani made 25% more than they were assigned, how many cookies were m
Marina CMI [18]

Answer:

80 cookies

Step-by-step explanation:

25% of 64 is 16
16+64=80

5 0
2 years ago
Need asapppppppp!!!!!
algol [13]

Answer:

122 degrees-

Step-by-step explanation:

m\angle {GOC}=m\angle{GOA}+m\angle{AOC}\\=32^\circ + 90^\circ\\=122^\circ

Angle GOA is a vertical angle with angle DOE, so it has the same measure (vertical angles are congruent).

Angle AOC is a right angle because angle COE is a right angle.

5 0
2 years ago
If you wish to warm 40 kg of water by 18 ∘C for your bath, find what the quantity of heat is needed. Express your answer in calo
Harman [31]

Answer:

Quantity of heat needed (Q) = 722.753 × 10³

Step-by-step explanation:

According to question,

Mass of water (m) = 40 kg

Change in temperature ( ΔT) = 18°c

specific heat capacity of water = 4200 j kg^-1 k^-1

The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .

So, Heat (Q) = m×s×ΔT

Or,          Q = 40 kg × 4200 × 18

or,           Q = 3024 × 10³ joule

Hence, Quantity of heat needed (Q) = 3024 × 10³ joule    

In calories 4.184 joule = 1 calorie

So, 3024 × 10³ joule = 722.753 × 10³

6 0
3 years ago
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