Question

1. communicate and justify your study partner says that the product of - 12m· - 3m is - 15m. what mistake did your study partner make?
2. augustina has 2 posters with length x inches. one poster has a width of x + 5 inches, and the other has a width of x + 7 inches. write an expression to represent the area of wall that the posters will cover.
3. multiply - 2.8x(3.7x + 8.1y).
4. what is the area of this rectangle?
5. use patterns and structure amoli says you should use the distributive property to multiply the two expressions. yuji says you should use the associative property. who is correct? explain.
6. higher order thinking can the expression 4x + 6x² be rewritten as 2x(2 + 3x)? explain.
7. select all the algebraic expressions equivalent to (5a + 6)4a. 18.ar.1.2

(11a)4a
20a²+24a
4a(5a + 6)
(-5a - 6)(-4a)
20a²

1. find the product. 18.ar.1.2

(1 + -\frac{3}{5}z)(-\frac{5}{3}z)
a -\frac{5}{3}z+z²
b \frac{5}{3}z+z²
c -\frac{2}{3}z - 2z
d 1+z²

Explanation:

Step1: Solve problem 22

The product of $- 12m\cdot - 3m$ should be calculated using the rule of multiplying two negative - numbers and the rule of multiplying variables with exponents. When multiplying two negative numbers, the result is positive, and when multiplying variables with the same base ($m$ in this case), we add the exponents (since $m = m^1$). So $-12m\cdot - 3m=(-12)\times(-3)\times m^{1 + 1}=36m^{2}$. The study - partner added the coefficients instead of multiplying them.

Step2: Solve problem 23

The area of the first poster with length $x$ inches and width $x + 5$ inches is $A_1=x(x + 5)=x^{2}+5x$ square inches. The area of the second poster with length $x$ inches and width $x + 7$ inches is $A_2=x(x + 7)=x^{2}+7x$ square inches. The total area $A$ that the two posters will cover is $A = A_1+A_2=(x^{2}+5x)+(x^{2}+7x)=2x^{2}+12x$ square inches.

Step3: Solve problem 24

Use the distributive property $a(b + c)=ab+ac$. Here, $a=-2.8x$, $b = 3.7x$, and $c = 8.1y$. So $-2.8x(3.7x + 8.1y)=-2.8x\times3.7x-2.8x\times8.1y=-10.36x^{2}-22.68xy$.

Step4: Solve problem 25

The area of a rectangle is given by $A = lw$, where $l=\frac{3}{2}b + 6$ and $w=\frac{2}{3}b$. Using the distributive property $A=\frac{2}{3}b(\frac{3}{2}b + 6)=\frac{2}{3}b\times\frac{3}{2}b+\frac{2}{3}b\times6=b^{2}+4b$ square feet.

Step5: Solve problem 26

The expression $(125 + y)\frac{1}{2}y$ should use the distributive property $a(b + c)=ab+ac$, where $a=\frac{1}{2}y$, $b = 125$, and $c = y$. So $(125 + y)\frac{1}{2}y=125\times\frac{1}{2}y+y\times\frac{1}{2}y=\frac{125}{2}y+\frac{1}{2}y^{2}$. Amoli is correct because the distributive property is used to multiply a single - term factor by a sum.

Step6: Solve problem 27

Use the distributive property to expand $2x(2 + 3x)$. We have $2x\times2+2x\times3x = 4x+6x^{2}$. So the expression $4x + 6x^{2}$ can be rewritten as $2x(2 + 3x)$.

Step7: Solve problem 28

Use the distributive property on $(5a + 6)4a=4a(5a + 6)=4a\times5a+4a\times6=20a^{2}+24a$. Also, $(5a + 6)4a$ is equivalent to $4a(5a + 6)$ and $(-5a - 6)(-4a)$ (since $(-5a - 6)(-4a)=(-1)(5a + 6)(-1)\times4a=(5a + 6)4a$). So the equivalent expressions are $20a^{2}+24a$, $4a(5a + 6)$, and $(-5a - 6)(-4a)$.

Step8: Solve problem 29

Use the distributive property on $(1-\frac{3}{5}z)(-\frac{5}{3}z)=1\times(-\frac{5}{3}z)-\frac{3}{5}z\times(-\frac{5}{3}z)=-\frac{5}{3}z + z^{2}$. So the answer is A.

Answer:

1. The study - partner added the coefficients instead of multiplying them. The correct product is $36m^{2}$.
2. $2x^{2}+12x$ square inches.
3. $-10.36x^{2}-22.68xy$.
4. $b^{2}+4b$ square feet.
5. Amoli is correct. We use the distributive property: $(125 + y)\frac{1}{2}y=\frac{125}{2}y+\frac{1}{2}y^{2}$.
6. Yes, because $2x(2 + 3x)=4x+6x^{2}$.
7. $20a^{2}+24a$, $4a(5a + 6)$, $(-5a - 6)(-4a)$
8. A. $-\frac{5}{3}z + z^{2}$