There are 10millimeters in one centimeter.
Since centimeter is a bigger measurement you'd multiply.
170 * 10 = 1700
There are 1700 millimeters in 170cm
Answer
1700 millimeters
<u>Step-by-step explanation:</u>
transform the parent graph of f(x) = ln x into f(x) = - ln (x - 4) by shifting the parent graph 4 units to the right and reflecting over the x-axis
(???, 0): 0 = - ln (x - 4)

0 = ln (x - 4)

1 = x - 4
<u> +4 </u> <u> +4 </u>
5 = x
(5, 0)
(???, 1): 1 = - ln (x - 4)

1 = ln (x - 4)

e = x - 4
<u> +4 </u> <u> +4 </u>
e + 4 = x
6.72 = x
(6.72, 1)
Domain: x - 4 > 0
<u> +4 </u> <u>+4 </u>
x > 4
(4, ∞)
Vertical asymptotes: there are no vertical asymptotes for the parent function and the transformation did not alter that
No vertical asymptotes
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transform the parent graph of f(x) = 3ˣ into f(x) = - 3ˣ⁺⁵ by shifting the parent graph 5 units to the left and reflecting over the x-axis
Domain: there is no restriction on x so domain is all real number
(-∞, ∞)
Range: there is a horizontal asymptote for the parent graph of y = 0 with range of y > 0. the transformation is a reflection over the x-axis so the horizontal asymptote is the same (y = 0) but the range changed to y < 0.
(-∞, 0)
Y-intercept is when x = 0:
f(x) = - 3ˣ⁺⁵
= - 3⁰⁺⁵
= - 3⁵
= -243
Horizontal Asymptote: y = 0 <em>(explanation above)</em>
Answer:
3 bulbs should be flawed
Step-by-step explanation:
2+ 18 = 20
20 bulbs were examined
2/20 = 1/10 were flawed
looking at the next 30 bulbs
30 * 1/10 = 3
3 of the next 30 bulbs should be flawed
Depends on which rate of change you're talking about. The rate of change is another term for a slope of a function. There's two(2) different version of rate of change.
First version one is the instantaneous rate of change. aka derivative. This one is found simply by taking the derivative of a function.
Second version is the average rate of change, which is found using the slope formula, (y₂ - y₁)/(x₂ - x₁)
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Initial value problem should give you an initial point (x, y) to plug into your function. You plug those x,y value in to find your answer.
There's variation of initial value problems so I can't give you any specific details on how to do it unless you can post the question.