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svet-max [94.6K]
3 years ago
10

Compare and contrast the results of adding or multiplying a value outside the argument of a function to the result of adding or

multiplying a value inside the argument of a function.
Mathematics
1 answer:
ki77a [65]3 years ago
5 0

Answer:

Only in special funcitons, specifically linear functions whose graphs pass through the origin, it is the same to put the constant inside the argument or outside

Step-by-step explanation:

If the function has the form f(x) = a*x, where a is a constant, then we have that f(c*x) = a*(c*x) = c*(a*x) = c*f(x). This kind of functions are proportional to the identity function f(x) = x, and they comprehend the linear functions whose graph pass through the origin.

For other linear function this property isnt true. For example if f(x) = x+4, then

f(4) = 4+4 = 8,

f(2*4) = 8+4 = 12

2*f(4) = 2*8 = 16

Thus, 2*f(4) is not f(2*4).

This property also isnt true for quadratic functions for example. If f(x) = x², then f(1) = 1, thus 3*f(1) = 3, however, f(3*1) = 3² = 9.

There might be coincidences for specific values, for example if f(x) = (x-1)*(x-2), then f(2*1)=2*f(1) = 0, however, for any other constant the result is not the same (for example f(0*1) = (-1)*(-2)=2, and 0*f(1) = 0).

If we want a function to satisfy the property f(c*x) = c*f(x) for any c,x, then it should be true that f(c) = f(c*1) = c*f(1). This means that if f(1) = a, then f(c) = c*a, so, in other words, f(x) = a*x = f(1) * x.

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A bag contains 42 red marbles, 6 white marbles, and 8 gray marbles. You randomly pick out a marble, record its color, and put it
Elanso [62]
For any single draw,

\mathbb P(\text{white})=\dfrac6{42+6+8}=\dfrac6{56}
\mathbb P(\text{gray})=\dfrac8{42+6+8}=\dfrac8{56}

Drawing a white marble or a gray marble are disjoint events; only one of them can happen. So

\mathbb P(\text{white or gray})=\mathbb P(\text{white})+\mathbb P(\text{gray})-\underbrace{\mathbb P(\text{white and gray})}_0
\mathbb P(\text{white or gray})=\dfrac6{56}+\dfrac8{56}=\dfrac{14}{56}

Out of 224 draws, you should expect \dfrac{14}{56}\times224=56 of the marbles to be either white or gray.
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4 years ago
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Solve for x: 12-2x=-2(y-x)
a_sh-v [17]
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Consider the vector field.f(x, y, z) = xy2z2i + x2yz2j + x2y2zk(a) find the curl of the vector field.
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First introduce the following notations:
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3 years ago
Suppose that the handedness of the last fifteen U.S. presidents is as follows:40% were left-handed (L)47% were Democrats (D)If a
lesantik [10]

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Probability that they were left handed P(L)= 40%

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So, P(L∩D) = 135 = 0.13

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marin [14]
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94.2 = \pi r^2\frac{10}3 \\ 282.6 = \pi r^2 *10 \\ 28.26 = \pi r^2 \\ 8.99543738355 = r^2 \\ 2.99923946752=r

Now we know that's going to be the radius of our <em>new </em>cone as well since we're keeping the diameter the same. The volume is going to be double 94.2 which is 188.4. Let's solve for the height.

188.4 = \pi (2.99923946752)^2\frac{h}3 \\ 188.4 = \pi(8.99543738355)\frac{h}3 \\ 188.4 = 28.26\frac{h}3 \\ 565.2 = 28.26h \\\\ \boxed{h = 20, r\approx 3}
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