All categories

3 years ago

Which of these is an example of the Commutative property of addition? And Explain why? 3+5=4+4 or 3+5=5+3

Mathematics

2 answers:

m_a_m_a [10]3 years ago

7 0

Answer:

3+5 = 5+3

Step-by-step explanation:

Communative Property states that the numbers on which we operate (the numbers you use to do addition and multiplication) can be moved or swapped from their position without making any difference to the answer. In your awnser it dosen't matter if you were to swich around the 5 and 3. You would still get the same awnser.

arlik [135]3 years ago

3 0

Answer:

3+5=5+3

Step-by-step explanation:

a+b=b+a

comunitive property

You might be interested in

Solve for X<br> 6(x + 2) = 66<br> X =

ioda

Answer:

6(x+2)=66

6x+12=66 -12

6x= 54 :6

x=9

Step-by-step explanation:

prove me wrong

4 0

3 years ago

2. The area of the rectangle shown is 375 square centimeters,

polet [3.4K]

Answer:

Step-by-step explanation:

Area is l x h. If that's the case then we need to do the oposite of multiplication to get the answer we need. Divide 375cm an 25cm to get your answer.

Original Equation:

A = l x h

New Equation:

A/l = h

8 0

3 years ago

Ricardo is building a new garden and is buying soil to fill it up. he buys 4 kilograms of soil for $10.00.

likoan [24]

Answer:

math duh

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Compute the following radical.<br> sq root 144

lara31 [8.8K]

we are given

$\sqrt{144}$

Firstly, we will find all possible factors of 144

$144=2\times 2\times 2\times 2\times 3\times 3$

we can also write as

$144=(2\times 2)^2\times (3)^2$

now, we can put exponent to over entire term

$144=(2\times 2\times 3)^2$

now, we can replace 144

$\sqrt{144}=\sqrt{(2\times 2\times 3)^2}$

now, sqrt will get cancelled with square

and we get

$\sqrt{144}=(2\times 2\times 3)$

now, we can simplify it

$\sqrt{144}=(4\times 3)$

$\sqrt{144}=12$ .............Answer

3 0

3 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

6 0

3 years ago