All categories

1 year ago

Name the property used in each step.

Mathematics

1 answer:

jarptica [38.1K]1 year ago

5 0

Answer:

$\begin{aligned}ab(a+b) & = (ab)a+(ab)b & & \textsf{Distributive Property of Addition} \\& = a(ab)+(ab)b & & \textsf{Commutative Property of Multiplication} \\& = (a \cdot a)b + a(b \cdot b) & & \textsf{Associative Property of Multiplication}\\& = a^2b+ab^2 & & \textsf{Property of Exponents}\end{aligned}$

Step-by-step explanation:

Distributive Property of Addition

Multiplying a number by a group of numbers added together is the same as multiplying each number separately.

Addition: a(b + c) = ab + ac

Subtraction: a(b - c) = ab – ac

Commutative Property

Changing the order or position of two numbers does not change the end result.

Applies to addition and multiplication only.

Addition: a + b = b + a

Multiplication: a × b = b × a

Associative Property

Grouping of numbers by parentheses in a different way does not affect their sum or product.

Applies to addition and multiplication only.

Addition: (a + b) + c = a + (b + c) = (a + c) + b

Multiplication: (a × b) × c = a × (b × c) = (a × c) × b

Property of Exponents

The exponent of a number shows how many times it should be multiplied by itself.

a × a × a = a³

b × b × b × b = b⁴

Learn more about Property Laws here:

brainly.com/question/28210762

You might be interested in

What is the equaiton of the line that goes through the point (6,-1) and is parallel to the line represented by the equautions y=

mixas84 [53]

parallel lines have the same slope

The slope-intercept form of a linear equatio is y=mx+b, where m stands for the "slope of the line" and b stands for the "y-intercept of the line"

They give you the equation y= -5/6x+3 Notice this is already on the slope-intercept form, so in this case the slope is -5/6 and the y-intercept is 3

You want an equation of the line that is parallel to the given line. The slopes must be the same, so m=-5/6

So far we have y=-5/6x + b

We don't have b yet but that can be found using the given point (6,-1) which tells you that "x is 6 when y is -1"

Replace that on the equation y=-5/6x + b and you get

-1 = (-5/6)(6) + b

-1 = -5 +b

4 = b

b = 4

We found b, or the y-intercept

Go back to the equation y = -5/6 x + b and replace this b with the b we just found

y = -5/6x + 4

6 0

2 years ago

124.4 divided by 4, idk they want more characters even tho it’s REALLY SIMPLE

klio [65]

Answer:

31.1

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Evaluate and simplify 4a - 2(a + b) + 1 = ? a=2 b=4

o-na [289]

Answer:

-3

Step-by-step explanation:

4(2)-2(2+4)+1=8-2(6)+1=9-12=-3

4 0

2 years ago

Read 2 more answers

Patrica is trying to compare the average rainfall of New York to that of Arizona. A comparison between these two states for the

Elanso [62]

Liters i guess. idk. im not rlly sure

7 0

3 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago