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nirvana33 [79]
4 years ago
12

The range of F(x) = logbx is the set of all positive real numbers.

Mathematics
1 answer:
shtirl [24]4 years ago
7 0

Answer:

The range of F(x) = logbx is the set of all positive real numbers is TRUE

Step-by-step explanation:

Given:

A function which logarithmic i.e F(x)=logbX=logx/logb  with base 10

To Find;

Range belongs to All set are positive real numbers.

Solution:

The domain is function for which all set of inputs are defined  and range for function is that set of all output that functions takes.

So Simple logarithmic function y=logbX is

X=b^y

So  The functions has domain of all real values  and range set of all real number.

In general the function F(x) = logbx  where X>0 and b≠1  is continuous and one to one function.

logarithmic function is not defined for negative numbers or for zero.

And Also function approaches y-axis as x-tends to infinity but never touches the it.

Hence the Given statement is true

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One of the interior angles of a triangle is equal to 30°, and one of the exterior angles is equal to 40°. Find the remaining int
irina1246 [14]

Answer:

The remaining interior angles of this triangle are 140º and 10º

Step-by-step explanation:

The sum of the interior angles of a triangle is always 180º.

A triangle has 3 angles. In this problem, we have one of them, that i am going to call A1 = 30º.

The sum of a interior angle with it's respective exterior angle is also always 180º.

We have that one of the exterior angles is equal to 40°. So it's respective interior angle is

40º + A2 = 180º

A2 = 180º - 40º

A2 = 140º

Now we have two interior angles, and we know that the sum of the 3 interior angles is 180º. So:

A1 + A2 + A3 = 180º

A3 = 180º - A1 - A2

A3 = 180º - 30º - 140º

A3 = 180º - 170º

A3 = 10º

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3 years ago
What do we call the statement that determines if the null hypothesis is rejected?
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Answer:

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Step-by-step explanation:

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3 years ago
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The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha
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Answer:

The average value of f over the interval [-2,5] is \frac{10}{7}.

Step-by-step explanation:

Let suppose that function f is continuous and integrable in the given intervals, by integral definition of average we have that:

\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6 (1)

\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20 (2)

By Fundamental Theorems of Calculus we expand both expressions:

\frac{F(3)-F(-2)}{3-(-2)} = -6

F(3) - F(-2) = -30 (1b)

\frac{F(5)-F(3)}{5-3} = 20

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We obtain the average value of f over the interval [-2, 5] by algebraic handling:

F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)

F(5) - F(-2) = 10

\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}

\bar f = \frac{10}{7}

The average value of f over the interval [-2,5] is \frac{10}{7}.

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