What is (2m+7)+(4d+6)
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(You go from left to right)
(Dividend is 931, divisor is 7)
Step 1: Look for a number smaller than 9 that is divisible by 7. 7 is the only one which is smaller than 9 and is divisible by 7. Now that you know 7*1=7, you put 1 above 931 (Dividend)
Step 2: You subtract that number from the number in tens place. And bring the leftover down.
Step 3: You repeat Step 1 and 2 until either you get a remainder of 0 or the remainder in smaller than the divisor (in this case 7)
(Therefore 931/7=133, quotient=133, remainder=0)
^Hope this helps.
(Dividend is 931, divisor is 7)
Step 1: Look for a number smaller than 9 that is divisible by 7. 7 is the only one which is smaller than 9 and is divisible by 7. Now that you know 7*1=7, you put 1 above 931 (Dividend)
Step 2: You subtract that number from the number in tens place. And bring the leftover down.
Step 3: You repeat Step 1 and 2 until either you get a remainder of 0 or the remainder in smaller than the divisor (in this case 7)
(Therefore 931/7=133, quotient=133, remainder=0)
^Hope this helps.
a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is
c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
<h3 /><h3>The domain of T</h3>Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
<h3>Inverse image of f(x)</h3>
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
<h3>Inverse image of g(x)</h3>
The inverse image of g(x) is
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
Replacing x with y, we have
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is
The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here: