51,000
5 in the ten thousands,
The thousands value is 1, the hundreds is 4 (less than 5 so we round down)
Answer:
x = 6
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define Equation</u>
3 - 2x = -1.5x
<u>Step 2: Solve for </u><em><u>x</u></em>
- Add 2x on both sides: 3 = 0.5x
- Divide 0.5 on both sides: 6 = x
- Rewrite: x = 6
<u>Step 3: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in<em> x</em>: 3 - 2(6) = -1.5(6)
- Multiply: 3 - 12 = -9
- Subtract: -9 = -9
Here we see that -9 does indeed equal -9.
∴ x = 6 is the solution to the equation.
Set like terms on each side by adding six to the first side and subtracting 1/2x from the first side. this gives you 9<1.5x. divide both sides by 1.5 to get 66. the answer is 4) x>6
The real solutions the equation as given in the task content; x² = 225 are; +25 and -25.
<h3>What are the real solutions of the equation as given in the task content?</h3>
It follows from the task content that the real solutions of the equation as given in the task content can be determined as follows;
x² = 225
x = ± 15
Therefore, the real solutions of the equation are; +25 and -25.
Read more on real solutions of equations;
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The time it will take the principal to grow to the desired amount is 0.7 years
Using the compound interest formula :
A = P(1 + r/n)^(nt)
A = final amount = 225,000
P = principal = 180,000
r = rate = 3.12%
n = Number of compounding times per period = 12(monthly)
t = time
225000 = 180000(1 + (0.0312 /12))^(12t)
Divide both sides by 180000
225000/180000 = (1 + (0.0312 /12))^(12t)
1.25 = 1.026^12t
Take the log of both sides
0.0969100 = 0.0111473 × 12t
0.0969100 = 0.1337676t
Divide both sides by 0.1337676 to isolate t
0.0969100 / 0.1337676 = t
0.7244 years
0.7 years
It will take 0.7 years for the amount to grow
Learn more : brainly.com/question/21270833?referrer=searchResults
