All categories

3 years ago

Questions is: the sum of 28+29+42=99. what addition property was used commutative, associative, identity, and distributive ?

Mathematics

1 answer:

Arturiano [62]3 years ago

8 0

Since, it is given that $28+29+42= 99$

Now, we have to identify the addition property for the given equation.

Associative property of addition states that for any real numbers say 'a' , 'b' and 'c', a+(b+c) = (a+b)+c

So, in the given question 28+29+42

We can use associative property as

(28+29)+42 = 28+(29+42)

57+42 = 28+71

99 = 99

Hence, among the given properties, associative addition property is used.

You might be interested in

Carlita has 2 1/2 pounds of turkey, the cherries are 1/4 equal to the turkey, and the chocolate drips are .35 equal to the cherr

denis23 [38]

Answer: There are 3.5 ounces of chocolate drips that Carlita need.

Step-by-step explanation:

Since we have given that

Number of pounds of turkey = $2\dfrac{1}{2}=\dfrac{5}{2}$

Since the cherries are $\dfrac{1}{4}$ equal to the turkey.

So, it becomes,

Number of pounds of cherries is given by

$\dfrac{1}{4}\times \dfrac{5}{2}\\\\=\dfrac{5}{8}\\\\=0.625$

Since the chocolate drips are 0.35 equal to the cherries.

So, the number of pounds of chocolate drips is given by

$0.625\times 0.35\\\\=0.21875\ lb$

Since 1 pound = 16 ounces.

So, Number of ounces of chocolate drips would be

$0.21875\times 16\\\\=3.5\ ounces$

Hence, there are 3.5 ounces of chocolate drips that Carlita need.

8 0

3 years ago

Two old RSM students were standing ten feet apart when they ran in opposite directions with speeds of 180 feet per minute and 16

inessss [21]

The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes

Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.

Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'

D = 180t + 10 + 160t

D = 340t + 10

<h3>Number of minutes before they are 1870 feet</h3>

Making t subject of the formula, we have

t = (D - 10)/340

Since they are 1870 feet apart after t minutes, D = 1870 feet.

t = (D - 10)/340

t = (1870 - 10)/340

t = 1860/340

t = 5.47 minutes

So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes