Login
Premium
Home Question Bank Math Algebra
Question
upstudy study bank question image url

Use the distributive property to remove the parentheses. \[ (5-5 x+3 u)(-2) \]

Ask by Smith Salazar. in the United States
Mar 14,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

\[ -10 + 10x - 6u \]

Solution

1. Begin with the expression: \[ (5 - 5x + 3u)(-2) \] 2. Distribute \(-2\) to each term inside the parentheses: \[ -2(5) + -2(-5x) + -2(3u) \] 3. Simplify each multiplication: \[ -10 + 10x - 6u \] 4. Therefore, the expression after applying the distributive property is: \[ -10 + 10x - 6u \]

Answered by UpStudy AI and reviewed by a Professional Tutor

error msg
Explain
Simplify this solution

Bonus Knowledge

To use the distributive property, you multiply each term inside the parentheses by -2. Here we go! First, distribute -2 to each term: \[ -2(5) + -2(-5x) + -2(3u) \] This results in: \[ -10 + 10x - 6u \] So, the expression after removing the parentheses is: \[ 10x - 6u - 10 \]

Related Questions

2.1 Calculate: \( 2.12 a(3 a-4 b)-3 b(b+2 a) \) \( 2.28 a^{2} \times-5 a \) (2) \( 2.3 \frac{2 x^{3} y^{2}}{4 x y} \) \( 2.4-20 a^{2}+b+5 a^{2}-3 b \) \( 2.5\left(a^{5} b^{2}\right)^{5} \) \( -2 \cdot 6-4^{4} \div 4^{-3} \)
Algebra South Africa Mar 16, 2025
7) For real numbers \( x, y, z \) such that \( x \neq y \neq z,\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0 \) and \( \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \neq 0 \) then \( x y z= \) \( \begin{array}{ll}\text { (A) } 2 & \text { (B) }-1 \\ \text { C) } 0 & \text { (D) } 1\end{array} \)
Algebra India Mar 16, 2025
\( x=4-\sqrt{15} \), find the value of \( \left(x+\frac{1}{x}\right) \)
Algebra India Mar 16, 2025
Quiz \( \frac{\text { Factor the expression completely }}{\text { (1) } 3 x^{3 / 2}-9 x^{\frac{1}{2}}+6 x^{-1 / 2}} \) (2) \( (2+x)^{-2 / 3} x+(2+x)^{1 / 3} \)
Algebra Kuwait Mar 16, 2025
\[ \text{Find } 2011^2 - 2010^2 \text{?} \]
Algebra South Africa Mar 16, 2025

Latest Algebra Questions

1) \( \left(3 x^{2}\right)\left(4 x^{3}\right) \) 2 Simplify simultar - \( \frac{10 x^{4}}{2 x^{2}} \) (1) A schao schoul bi pnce of 3 Simplify (2) \[ (x+2)(x-3) \] the 4 Simply Simplify \[ \frac{6 x^{3} y^{2}}{2 x^{2} y} \] (3) 5 untiply: \[ (2 x+3)(x-4) \] © 6 Multiply \[ 4(x+2 y-5) \] Simplify \[ \left(x^{2}-4 x+4\right) \div(x-2) \] Multiply and Simplify \[ (x+1)\left(x^{2}-x+3\right) \] Find the product: \[ (x-5)(x+5) \]
Algebra Papua New Guinea Mar 16, 2025
2.1 Calculate: \( 2.12 a(3 a-4 b)-3 b(b+2 a) \) \( 2.28 a^{2} \times-5 a \) (2) \( 2.3 \frac{2 x^{3} y^{2}}{4 x y} \) \( 2.4-20 a^{2}+b+5 a^{2}-3 b \) \( 2.5\left(a^{5} b^{2}\right)^{5} \) \( -2 \cdot 6-4^{4} \div 4^{-3} \)
Algebra South Africa Mar 16, 2025
\[ \text{Find } 2011^2 - 2010^2 \text{?} \]
Algebra South Africa Mar 16, 2025
7) For real numbers \( x, y, z \) such that \( x \neq y \neq z,\left|\begin{array}{lll}x & x^{2} & 1+x^{3} \\ y & y^{2} & 1+y^{3} \\ z & z^{2} & 1+z^{3}\end{array}\right|=0 \) and \( \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| \neq 0 \) then \( x y z= \) \( \begin{array}{ll}\text { (A) } 2 & \text { (B) }-1 \\ \text { C) } 0 & \text { (D) } 1\end{array} \)
Algebra India Mar 16, 2025
B) \( \frac{7}{13} \times \frac{\square}{\square}=\frac{7}{13} \)
Algebra Mexico Mar 16, 2025
Prev Next
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy