/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 $$4 \times 5$$... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 29

$$4 \times 5$$

Short Answer

Expert verified
The result of multiplying 4 by 5 is 20.

Step by step solution

01

Understand the Problem

The problem given is a multiplication problem, where we need to multiply 4 by 5.
02

Recall Multiplication Concept

Multiplication is a way of adding the same number multiple times. Here, it means adding the number 4, five times.
03

Perform the Multiplication

We perform the multiplication by either recalling the multiplication table or adding as follows: \(4 + 4 + 4 + 4 + 4 = 20\).
04

Verify Your Answer

Check the answer by reversing the multiplication, \(5 imes 4\), which also equals 20. This verifies our solution is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Multiplication Table
The multiplication table is a mathematical tool often introduced in early schooling. It provides a quick and reliable way to find the result of a multiplication problem like the one given: \(4 \times 5\). Each cell in the table corresponds to a unique product of two numbers.
For instance, to find \(4 \times 5\) with a multiplication table, locate the row for 4 and the column for 5. The intersecting cell shows the product, which is 20. Using multiplication tables helps:
  • Improve speed and accuracy
  • Build a strong foundational understanding of numbers
  • Serve as a quick reference guide
With regular practice, you can easily recall products and develop mental math skills efficiently.
Repeated Addition
Repeated addition is an intuitive approach to understanding multiplication. It involves adding the same number multiple times. This method is often used to explain the concept of multiplication to beginners. In the case of \(4 \times 5\), you would add the number 4, five times, like this:
\[ 4 + 4 + 4 + 4 + 4 = 20 \]
This method emphasizes:
  • The connection between addition and multiplication
  • Visualization of mathematical operations
  • A procedural understanding of how numbers are grouped

Repeated addition lays the groundwork for grasping more complex multiplication and helps in connecting arithmetic operations logically.
Verification of Solutions
Verifying your multiplication result ensures accuracy and confidence in your answer. Suppose you've calculated \(4 \times 5\) and got 20. To verify, you can reverse the operation, swapping the numbers: \(5 \times 4\), which also equals 20. This check confirms that the order of multiplication doesn't affect the product due to the commutative property of multiplication:
\[ a \times b = b \times a \]
Verification helps to:
  • Build confidence in problem-solving skills
  • Avoid careless mistakes
  • Understand properties like commutativity and associativity
Remember, checking your work is a valuable habit that can apply to more advanced math concepts and everyday math-related tasks.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

We solve \(\operatorname{det}(\mathbf{A}-\lambda \mathbf{I})=\left|\begin{array}{ccc}-\lambda & 4 & 0 \\ -1 & -4-\lambda & 0 \\\ 0 & 0 & -2-\lambda\end{array}\right|=-(\lambda+2)^{3}=0\). For \(\lambda_{1}=\lambda_{2}=\lambda_{3}=-2\) we have \(\left(\begin{array}{rrr|r}2 & 4 & 0 & 0 \\ -1 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right) \Rightarrow\left(\begin{array}{lll|l}1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{array}\right)\) so that \(k_{1}=-2 k_{2}\). If \(k_{2}=1\) and \(k_{3}=1\) then \(\mathbf{K}_{1}=\left(\begin{array}{r}-2 \\ 1 \\\ 0\end{array}\right) \quad\) and \(\quad \mathbf{K}_{2}=\left(\begin{array}{l}0 \\\ 0 \\ 1\end{array}\right)\).

The decoded message is $$\mathbf{M}=\mathbf{A}^{-1} \mathbf{B}=\left(\begin{array}{rr} 2 & -3 \\ -5 & 8 \end{array}\right)\left(\begin{array}{ccccc} 152 & 184 & 171 & 86 & 212 \\ 95 & 116 & 107 & 56 & 133 \end{array}\right)=\left(\begin{array}{ccccc} 19 & 20 & 21 & 4 & 25 \\ 0 & 8 & 1 & 18 & 4 \end{array}\right)$$ From correspondence (1) we obtain: STUDY_HARD.

The system of equations is $$\begin{aligned} i_{1}-i_{2}-i_{3} &=0 & & i_{1}-i_{2}-i_{3}=0 \\ 52-i_{1}-5 i_{2} &=0 & \text { or } \quad & i_{1}+5 i_{2} =52 \\ -10 i_{3}+5 i_{2} &=0 & & 5 i_{2}-10 i_{3} =0 \end{aligned}$$ Gaussian elimination gives $$\left(\begin{array}{rrr|r} 1 & -1 & -1 & 0 \\ 1 & 5 & 0 & 52 \\ 0 & 5 & -10 & 0 \end{array}\right) \frac{\operatorname{row}}{\text { operations }}\left(\begin{array}{rrr|c} 1 & -1 & -1 & 0 \\ 0 & 1 & 1 / 6 & 26 / 3 \\ 0 & 0 & 1 & 4 \end{array}\right).$$ The solution is \(i_{1}=12, i_{2}=8, i_{3}=4\).

(a) We write the equations in the form \\[ \begin{aligned} -4 u_{1}+u_{2}+u_{4} &=-200 \\ u_{1}-4 u_{2}+u_{3} &=-300 \\ u_{2}-4 u_{3}+u_{4} &=-300 \\ u_{1}+u_{3}-4 u_{4} &=-200 \end{aligned} \\] In matrix form this becomes \(\left(\begin{array}{rrrr}-4 & 1 & 0 & 1 \\ 1 & -4 & 1 & 0 \\ 0 & 1 & -4 & 1 \\ 1 & 0 & 1 & -4\end{array}\right)\left(\begin{array}{l}u_{1} \\ u_{2} \\ u_{3} \\\ u_{4}\end{array}\right)=\left(\begin{array}{c}-200 \\ -300 \\ -300 \\\ -200\end{array}\right).\) (b) \(\mathbf{A}^{-1}=\left(\begin{array}{llll}-\frac{7}{24} & -\frac{1}{12} & -\frac{1}{24} & -\frac{1}{12} \\ -\frac{1}{12} & -\frac{7}{24} & -\frac{1}{12} & -\frac{1}{24} \\ -\frac{1}{24} & -\frac{1}{12} & -\frac{7}{24} & -\frac{1}{12} \\ -\frac{1}{12} & -\frac{1}{24} & -\frac{1}{12} & -\frac{7}{24}\end{array}\right) ; \quad \mathbf{A}^{-1}\left(\begin{array}{c}-200 \\ -300 \\ -300 \\\ -200\end{array}\right)=\left(\begin{array}{c}\frac{225}{2} \\ \frac{275}{2} \\\ \frac{275}{2} \\ \frac{225}{2}\end{array}\right) ; \quad u_{1}=u_{4}=\frac{225}{2}, u_{2}=u_{3}=\frac{275}{2}\)

From \(x_{1} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+x_{2} \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow x_{3} \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}\) we obtain the system $$3 x_{1}=x_{3}, \quad 2 x_{1}+x_{2}=2 x_{3}, \quad 8 x_{1}+4 x_{2}=8 x_{3}, \quad 3 x_{2}=4 x_{3}.$$ Letting \(x_{1}=t\) we see from the first equation that \(x_{3}=3 t\) and from the fourth equation that \(x_{2}=4 t .\) These choices also satisfy the second and third equations. Taking \(t=1\) we obtain the balanced equation $$\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+4 \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 3 \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}.$$
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Solid State Physics

Read Explanation

Radiation

Read Explanation

Fields in Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.