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2 years ago

Can the exponent in the denominator be subtracted from the exponent in the numerator when the based are different? Explain

Mathematics

2 answers:

cricket20 [7]2 years ago

6 0

Answer:no

Step-by-step explanation:

Because u can not subtract a denominator from numerator u must subtract the denominator by the denominator

erastova [34]2 years ago

5 0

Answer:

If the higher power is in the denominator, put the difference in the denominator and vice versa, this will help avoid negative exponents and a repeat of step 3. Step 6: Raise each coefficient (or number) to the appropriate power and then simplify or reduce any remaining fractions. Step 1: Apply the Zero-Exponent Rule.

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3 years ago

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How do you show where an inequality is true on a number line?

Mashutka [201]

Answer:

When the two values are different

Step-by-step explanation:

Like in an inequality equation.

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$4+5x\geq 30$

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6 0

2 years ago

Simplify 6 - 3(4 - 3x)<br> -3x - 6<br> 9x + 18<br> 9x - 6

svet-max [94.6K]

Answer:

6 (19 x - 2)

Step-by-step explanation:

Simplify the following:

6 - 3 (4 - 3 x) - 3 x - 6×9 x + 18×9 x - 6

-6×9 = -54:

6 - 3 (4 - 3 x) - 3 x + -54 x + 18×9 x - 6

18×9 = 162:

6 - 3 (4 - 3 x) - 3 x - 54 x + 162 x - 6

Grouping like terms, 6 - 3 (4 - 3 x) - 3 x - 54 x + 162 x - 6 = (-3 x - 54 x + 162 x) - 3 (4 - 3 x) + (6 - 6):

(-3 x - 54 x + 162 x) - 3 (4 - 3 x) + (6 - 6)

-3 x - 54 x + 162 x = 105 x:

105 x - 3 (4 - 3 x) + (6 - 6)

6 - 6 = 0:

105 x - 3 (4 - 3 x)

-3 (4 - 3 x) = 9 x - 12:

9 x - 12 + 105 x

Grouping like terms, 105 x + 9 x - 12 = (105 x + 9 x) - 12:

(105 x + 9 x) - 12

105 x + 9 x = 114 x:

114 x - 12

Factor 6 out of 114 x - 12:

Answer: 6 (19 x - 2)

7 0

2 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = 3 cos2(x) − 6 sin(x), 0 ≤ x ≤ 2π (a) Find the inte

marta [7]

Answer:

f(x) = 3 cos2(x) − 6 sin(x), 0 ≤ x ≤ 2π

We plot the function using a graphing calculator (See attached image below)

a) 1. Find the interval on which f is increasing.

f(x) increases for

x ∈ [π/2 , 7π/6] ∪ [3π/2, 11π/6]

2. Find the interval on which f is decreasing

f(x) decreases for

x ∈ [0 , π/2) ∪ (7π/6, 3π/2) ∪ (11π/6, 2π]

(b) Find the local minimum and maximum values of f

Local minimum. f = -9

Local maximum. f = 4.5

Please see second image attached

Points,

(x,y) = (2.50673, -2.66882)

(x,y) = (4.14456, 3.79382)

(x,y) = (5.28022, 3.79382)

2. Find the interval on which f is concave up.

x ∈ [0 , 2.50673] ∪ [4.14456, 5.28022]

3. Find the interval on which f is concave down.

x ∈ (2.50673, 4.14456) ∪ (5.28022, 2π]

7 0

2 years ago