Question

1. all the values in the equation are exact. what value of c makes the equation true?

(4.25×10^{2c + 4})(6×10^{7}) = 255
f. -7
g. -6.5
h. -5
j. -4.5

1. which of the following inequalities is true for all positive integers m?

a. m ≤ 1/m
b. m ≤ √m
c. m ≥ m²
d. m ≤ m + 1

1. graphed in the standard (x,y) coordinate plane is a right triangle with vertices (0,0), (-40,0), and (0,30). what is the length, in coordinate units, of the hypotenuse of the triangle?

f. 30
g. 40
h. 50
j. 70

Explanation:

Problem 22

Step1: Multiply the coefficients

Multiply \(4.25\) and \(6\): \(4.25\times6 = 25.5\)

Step2: Use the rule of exponents \(a^m\times a^n=a^{m + n}\)

For the powers of \(10\), we have \(10^{2c + 4}\times10^{7}=10^{2c+4 + 7}=10^{2c + 11}\)
So, \((4.25\times10^{2c + 4})(6\times10^{7})=25.5\times10^{2c+11}\)

Step3: Rewrite \(25.5\times10^{2c + 11}\) in standard form

\(25.5\times10^{2c+11}=2.55\times10^{1}\times10^{2c + 11}=2.55\times10^{2c+12}\)
Since \((4.25\times10^{2c + 4})(6\times10^{7}) = 255=2.55\times10^{2}\)

Step4: Set the exponents equal

Set \(2c+12 = 2\)
Subtract \(12\) from both sides: \(2c=2 - 12=-10\)
Divide both sides by \(2\): \(c=-5\)

• For option A: If \(m = 2\), then \(m=2\) and \(\frac{1}{m}=\frac{1}{2}\), and \(2>\frac{1}{2}\), so \(m\leq\frac{1}{m}\) is false for \(m = 2\).
• For option B: If \(m = 4\), then \(m = 4\) and \(\sqrt{m}=2\), and \(4>2\), so \(m\leq\sqrt{m}\) is false for \(m = 4\).
• For option C: If \(m = 2\), then \(m = 2\) and \(m^{2}=4\), and \(2<4\), so \(m\geq m^{2}\) is false for \(m = 2\).
• For option D: Subtract \(m\) from both sides of \(m\leq m + 1\), we get \(0\leq1\), which is True for all positive integers \(m\).

Step1: Use the distance formula for the length of the sides

The length of one leg \(a\) (along the \(x\) - axis) with endpoints \((0,0)\) and \((-40,0)\) is \(| - 40-0|=40\)
The length of the other leg \(b\) (along the \(y\) - axis) with endpoints \((0,0)\) and \((0,30)\) is \(|30 - 0|=30\)

Step2: Use the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\)

Here \(a = 40\) and \(b = 30\), then \(c=\sqrt{40^{2}+30^{2}}=\sqrt{1600 + 900}=\sqrt{2500}=50\)

Answer:

H. \(-5\)

Problem 23