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Chapter 3: Problem 4

Solve the equations. \(0.32=\frac{2.56}{\sqrt{y}}\)

Short Answer

Expert verified
y = 64

Step by step solution

01

Isolate the Square Root

First, multiply both sides of the equation by \(\sqrt{y}\) to isolate the square root term on one side of the equation: \[ 0.32 \sqrt{y} = 2.56 \]
02

Solve for the Square Root

Next, divide both sides by 0.32 to solve for \(\sqrt{y}\): \[ \sqrt{y} = \frac{2.56}{0.32} \]
03

Simplify the Right Side

Calculate the value of the right-hand side: \[ \frac{2.56}{0.32} = 8 \] So, \(\sqrt{y} = 8\).
04

Solve for y

Square both sides of the equation to solve for \(y\): \[ y = 8^2 = 64 \]

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Key Concepts

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

square root isolation
When you encounter an equation with a square root, the first step is usually to isolate the square root on one side of the equation. This helps to simplify the overall problem. In this specific example given, we have the equation:
0.32 = \( \frac{2.56}{\sqrt{y}}\).
To start isolating the square root, multiply both sides by \( \sqrt{y} \), thus eliminating the fraction. This results in: 0.32 \( \sqrt{y} = 2.56 \).
Isolating the square root in this manner sets you up for the next steps to solve the equation.
simplifying fractions
Simplifying fractions is a core skill in algebra, as it makes calculations cleaner and easier. After isolating the square root, our next step involves simplifying the fraction on the right-hand side of the equation:
<\( \sqrt{y} = \frac{2.56}{0.32} \).
To do this, perform the division: <\( \frac{2.56}{0.32} = 8 \).
This simplification means we now have <\( \sqrt{y} = 8 \), which drastically simplifies our equation and moves us closer to solving for y.
exponents
Dealing with exponents is another crucial part of solving equations involving square roots. Once we have \( \sqrt{y} = 8 \), we need to get rid of the square root to solve for y. The efficient way to do this is by squaring both sides of the equation.
Recall that squaring a square root effectively cancels out the square root.
Thus, we get: <\( (\sqrt{y})^2 = 8^2 \), which simplifies to y = <\( 64 \).
Understanding this manipulation of exponents is essential for solving many algebraic problems.
algebraic manipulation
Algebraic manipulation involves rearranging and simplifying expressions to solve for unknown variables. In this problem, we used several key manipulations:
  • First, isolating the square root by multiplying both sides by \( \sqrt{y} \).

  • Second, simplifying the fraction by performing the division.

  • Third, eliminating the square root by squaring both sides of the equation.
Understanding and practicing these algebraic manipulation techniques will help you solve similar equations with confidence.

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Most popular questions from this chapter

Suppose that \(y\) varies jointly as \(x^{4}\) and \(w\). If \(x\) is replaced by \(\frac{1}{4} x\) and \(w\) is replaced by \(4 w,\) what is the effect on \(y ?\)

a. Factor the polynomial over the set of real numbers. b. Factor the polynomial over the set of complex numbers. $$f(x)=x^{4}-6 x^{3}+9 x^{2}-6 x+8$$

The front face of a tent is triangular and the height of the triangle is two- thirds of the base. The length of the tent is \(3 \mathrm{ft}\) more than the base of the triangular face. If the tent holds a volume of \(108 \mathrm{ft}^{3}\), determine its dimensions.

Determine if the statement is true or false. If \(b\) is an upper bound for the real zeros of a polynomial, then \(-b\) is a lower bound for the real zeros of the polynomial.

a. Factor the polynomial over the set of real numbers. b. Factor the polynomial over the set of complex numbers. $$f(x)=x^{4}+2 x^{3}+x^{2}+8 x-12$$
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