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andrew-mc [135]
3 years ago
10

Identify the equivalent expression for (m^1/3 m^1/5)^0

Mathematics
2 answers:
Rama09 [41]3 years ago
6 0

Answer:

1. is the answer

Step-by-step explanation:

Lostsunrise [7]3 years ago
4 0

Answer:

1.

Step-by-step explanation:

The thing is, x^0 is always 1. So even if you simplify, it will still be 1.

(m^\frac{1}{3}*m^\frac{1}{5})^0\\(m^\frac{8}{15})^0\\\text{But when put to the power of 0,}\\1.

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Find the Quotient and Remainder. 5x^3 - 6x^2 + x + 7 divided by x^2
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Step-by-step explanation:

Solution

According to remainder theorem, when f(x) is divided by (x+2), Remainder =f(−2)

f(x)=5x3+2x2−6x+12

f(−2)=5(−2)3+2(−2)2−6(−2)+12

=5×−8+2×4+12+12

=−40+32=−8

∴ Remainder=−8

5 0
3 years ago
The difference of d minus 3 divided by 2
Eduardwww [97]

Answer:

(d-3)/2

Step-by-step explanation:

Difference is subtraction

(d-3)/2

6 0
3 years ago
If f(x) = 2x + 3 and g(x) =4x - 1, find f(4).
fomenos

Solve the "f" function with substitute 4 and solve the "g" function with what we get for the "f" function.

f(4) = 2(8) + 3

f(4) = 16 + 3

f(4) = 19

g(19) = 4(19) - 1

g(19) = 76 - 1

g(19) = 75

Best of Luck!

6 0
3 years ago
What is thisssssssss
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Answer:

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Step-by-step explanation:

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3 years ago
What is the equation of the function shown in the graph provided?<br><br> y= _ x + _
Mkey [24]

Answer:

y = -3x - 6

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Order of Operations: BPEMDAS

<u>Algebra I</u>

Slope Formula: m=\frac{y_2-y_1}{x_2-x_1}

Slope-Intercept Form: y = mx + b

  • m - slope
  • b - y-intercept

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Find points from graph.</em>

x-intercept (-2, 0)

y-intercept (0, -6)

<u>Step 2: Find slope </u><em><u>m</u></em>

  1. Substitute:                  m=\frac{-6-0}{0+2}
  2. Subtract/Add:             m=\frac{-6}{2}
  3. Divide:                        m=-3

<u>Step 3: Redefine</u>

Slope <em>m</em> = -3

y-intercept <em>b</em> = -6

<u>Step 4: Write linear equation</u>

Slope-Intercept Form:   y = -3x - 6

7 0
3 years ago
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