All categories

3 years ago

A function is a relation where each input value is assigned to___ one output value.

Mathematics

2 answers:

astra-53 [7]3 years ago

5 0

Answer:

The correct answers with explanation are as follows.

Step-by-step explanation:

1. A function is a relation where each input value is assigned to ONLY one output value. When each input value has one and only one output value, that relation is a function.

2. The domain of a function is the set of all input values, or x-values, for which the function is defined. The domain is the input or the independent value, that goes into a function. And the output that comes out is the range.

3. The range of a function is the set of all output values, or y-values, for which the function is defined. As mentioned above in the 2nd point, the range is the output or dependent value, that comes after the input has been processed.

4. To write the equation y = ax + b in function notation, substitute f(x) for y. this is called function of 'x'. And it is solved by substituting values for 'x' in ax+b.

irakobra [83]3 years ago

3 0

A function is a relation where each input value is assigned to only one output value.

The domain of a function is the set of all input values, or x-values, for which the function is defined.

The range of a function is the set of all output values, or y-values, for which the function is defined.

To write the equation y = ax + b in function notation, substitute f(x) for y.

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Step-by-step explanation:

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3 years ago

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3 years ago

Divide 50 into the ratio 1:4

Alex787 [66]

The answer is 10:40
so 1:4
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7 0

3 years ago

Read 2 more answers

Does anyone know how to do this ?

goldenfox [79]

Answer: D) 32

Step-by-step explanation:

The area of a parallelogram equals its base times its height. Substitute the base and area into this formula to solve for the height:

A=bh

40=(5)(h)

h=8

If ABCD is a parallelogram, opposite sides are parallel, so AB is parallel to DC. If AB is parallel to DC, EB is parallel to DC. Since EB is parallel to DC, quadrilateral EBCD has at least one pair of parallel sides, making it a trapezoid. The area of a trapezoid is equal to $\frac{h(b_{1}+b_{2}) }{2}$ where h is the height, $b_{1}$ is one base, and $b_{2}$ is the other base. Plug the necessary values into the formula:

A= $\frac{h(b_{1}+b_{2}) }{2}$

$b_{1} =AB-AE\\DC=AB\\b_{1} =DC-AE\\b_{1} =5-2\\b_{1} =3$

$b_{2} =5$

h=8

A= $\frac{8(3+5)}{2}$

A=32

The answer is D) 32.

6 0

3 years ago

Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

$\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}$

$\frak {\huge{ \frac{1}{1 + {x}^{2} } }}$

Step-by-step explanation:

$\sf let \: f(x) = { \tan }^{ - 1} x \\ \\ \sf f(x + h) = { \tan}^{ - 1} (x + h)$

$\sf f'(x) = \frac{f(x+h) - f(x) }{h}$

$\sf \implies \lim_{ h \to 0 } \frac{ { \tan }^{ - 1}(x + h) - { \tan}^{ - 1}x }{h} \\ \\ \\ \sf \implies \lim_ {h \to 0} \frac{ { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}$

By using

$\sf { \tan}^{ - 1} x - { \tan}^{ - 1} y = \\ \sf { \tan}^{ - 1} \frac{x - y}{1 + xy} formula$

$\sf \implies \large \lim_{h \to0 } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{h} \\ \\ \\ \sf \implies \large{\lim_{h \to0} } \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } \times (1 + hx + {x}^{2} )} \\ \\ \\ \sf \implies \large \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx + {x}^{2} } }{ \frac{h}{1 + hx + {x}^{2} } } + \lim_{h \to0} \frac{1}{1 + hx + {x}^{2} }$

Now putting the value of h = 0

$\sf \large \implies 0 + \frac{1}{1 + 0 + {x}^{2} } \\ \\ \\ \purple{ \boxed { \implies \frac{1}{1 + {x}^{2} } }}$

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2 years ago