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2-6 additional practice leveled practice in 1 - 4, apply the properties…

Question

2-6 additional practice
leveled practice in 1 - 4, apply the properties of operations to multiply the expressions.

  1. 2.4x·(−5x)

=2.4·5·x·x
=
x

  1. (y - 1/2)(-3/4y)

=(y)( )+( )(-3/4y)
=-3/4y + 3/8

  1. (1.5 + 2.3m)(-4.2m)

=(1.5 + 2.3m)
=-9.66m²

  1. -1/5n(2n - 1)

=n
n
in 5 - 10, use the properties of operations to multiply the expressions.

  1. (-1/10x)(3/5x)
  2. (2 + 2/3y)(1/2y)
  3. (5.3a)(a - 7.8)
  4. (-2.6b)(3.8 - b)
  5. (k + k)(8.9k)
  6. (7 - 1/4p - 1)(-5/6p)
  7. what is the area of this rectangle?
  8. communicate and justify jose said the product of -1.4x·(-4x) is 2.6x. what mistake did he make?

b + 1.6 in.
12.16 in.

Explanation:

Step1: Multiply coefficients and variables

For \(2.4x\cdot(- 5x)\), we multiply the coefficients \(2.4\) and \(-5\), and the variables \(x\) and \(x\).
\[2.4\times(-5)\times x\times x=-12x^{2}\]

Step2: Expand \((y - \frac{1}{2})(-\frac{3}{4}y)\) using distributive property

\[y\times(-\frac{3}{4}y)-\frac{1}{2}\times(-\frac{3}{4}y)=-\frac{3}{4}y^{2}+\frac{3}{8}y\]

Step3: Expand \((1.5 + 2.3m)(-4.2m)\) using distributive property

\[1.5\times(-4.2m)+2.3m\times(-4.2m)=-6.3m-9.66m^{2}\]

Step4: Expand \(-\frac{1}{5}n(2n - 1)\) using distributive property

\[-\frac{1}{5}n\times2n-(-\frac{1}{5}n\times1)=-\frac{2}{5}n^{2}+\frac{1}{5}n\]

Step5: Multiply \((-\frac{1}{10}x)(\frac{3}{5}x)\)

\[(-\frac{1}{10})\times\frac{3}{5}\times x\times x=-\frac{3}{50}x^{2}\]

Step6: Expand \((2+\frac{2}{3}y)(\frac{1}{2}y)\) using distributive property

\[2\times\frac{1}{2}y+\frac{2}{3}y\times\frac{1}{2}y=y + \frac{1}{3}y^{2}\]

Step7: Expand \((5.3a)(a - 7.8)\) using distributive property

\[5.3a\times a-5.3a\times7.8=5.3a^{2}-41.34a\]

Step8: Expand \((-2.6b)(3.8 - b)\) using distributive property

\[(-2.6b)\times3.8-(-2.6b)\times b=-9.88b + 2.6b^{2}\]

Step9: Simplify \((k + k)(8.9k)\)

First, \(k + k = 2k\), then \(2k\times8.9k=17.8k^{2}\)

Step10: Expand \((7-\frac{1}{4}p - 1)(-\frac{5}{6}p)\)

Simplify \(7-\frac{1}{4}p - 1=6-\frac{1}{4}p\), then \((6-\frac{1}{4}p)(-\frac{5}{6}p)=6\times(-\frac{5}{6}p)-\frac{1}{4}p\times(-\frac{5}{6}p)=-5p+\frac{5}{24}p^{2}\)

Step11: Find area of rectangle

The area of a rectangle with length \(l = 12.1b\) inches and width \(w=b + 1.6\) inches is \(A=l\times w=12.1b(b + 1.6)=12.1b^{2}+19.36b\) square - inches

Step12: Analyze Jose's mistake

For \(-1.4x\cdot(-4x)\), we multiply the coefficients \(-1.4\) and \(-4\) to get \(5.6\), and \(x\cdot x=x^{2}\), so the product is \(5.6x^{2}\). Jose made a mistake in multiplying the coefficients \(-1.4\) and \(-4\) incorrectly.

Answer:

  1. \(-12x^{2}\)
  2. \(-\frac{3}{4}y^{2}+\frac{3}{8}y\)
  3. \(-6.3m - 9.66m^{2}\)
  4. \(-\frac{2}{5}n^{2}+\frac{1}{5}n\)
  5. \(-\frac{3}{50}x^{2}\)
  6. \(y+\frac{1}{3}y^{2}\)
  7. \(5.3a^{2}-41.34a\)
  8. \(-9.88b + 2.6b^{2}\)
  9. \(17.8k^{2}\)
  10. \(-5p+\frac{5}{24}p^{2}\)
  11. \(12.1b^{2}+19.36b\) square - inches
  12. Jose multiplied the coefficients \(-1.4\) and \(-4\) incorrectly; the correct product is \(5.6x^{2}\)