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olga nikolaevna [1]
2 years ago
7

If x² = 64, then 1/3 + x⁰​

Mathematics
2 answers:
Alenkasestr [34]2 years ago
8 0

Answer:

\huge\boxed{\bf\:\frac{4}{3} }

Step-by-step explanation:

If, x^{2} = 64, then,

x^{2} = 64\\x = \sqrt{64}\\x = (+/-)8

Now, remember that any number when folllowed by exponent 0 will always have its value as 1.

Then,

\frac{1}{3} + x^{0} \\ =\frac{1}{3} + (+/-8)^{0} \\= \frac{1}{3} + 1\\=\frac{1}{3} +\frac{3}{3} \:\:\:\: (LCM = 3)\\ \boxed{\bf\:\frac{4}{3} }

\rule{150pt}{2pt}

Elodia [21]2 years ago
5 0

Answer:

<u>4/3 or 1 1/3</u>

Step-by-step explanation:

<u>Finding x</u>

  • x² = 64
  • x = ±8

<u>Substitute in the equation</u>

  • 1/3 + (±8)⁰
  • 1/3 + 1
  • <u>4/3 or 1 1/3</u>
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Answer:

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Step-by-step explanation:

A linear equation is an equation in the form y = mx + b, where m is the slope of the line and b is the y intercept.

Equation A is represented by the points:

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Equation A can be gotten by picking any two pair of points. Let us use (-5, -4) and (0, 1). We use this formula:

y- y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-(-4)=\frac{1-(-4)}{0-(-5)} (x-(-5))\\\\y +4=x+5\\\\y=x+1

Equation B is represented by the points:

x:    -6      -3        3          6

y:    -4       -2         2          4

Equation B can be gotten by picking any two pair of points. Let us use (-6, -4) and (3, 2). We use this formula:

y- y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\y-(-4)=\frac{2-(-4)}{3-(-5)} (x-(-6))\\\\y +4=\frac{2}{3}( x+6)\\\\y+4=\frac{2}{3} x+4\\\\y=\frac{2}{3}x

The solution to the system of equations is gotten by solving y = x + 1 and y = (2/3)x simultaneously.

Substitute y = (2/3)x in y = x + 1:

(2/3)x = x + 1

2x = 3x + 3

x = -3

Put x = -3 in y = (2/3)x

y = (2/3) (-3) = -2

Hence (-3, -2) is the solution

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5. Prove the conditional by proving the contrapositive:
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Answer + Step-by-step explanation:

<u>the conditional statement</u> :

For two positive integers n and m,

nm = 4 ⇒  (n = 1  or  m = 1)  or  (n=m= 2)

<u>the contrapositive</u>:

(n ≠ 1  and  m ≠ 1)  and  (n ≠ 2 or m ≠ 2) ⇒ nm ≠ 4

<u>Prove the conditional by proving the contrapositive</u> :

Suppose

m ≠ 1

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<u>Conclusion</u>:

The contrapositive is true then conditional is true.

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