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3 years ago

Find the range of the function y=1/2x+3 when the domain is (-2,0,2,4)

Mathematics

1 answer:

Triss [41]3 years ago

5 0

Answer:

{2,3,4,5}

Step-by-step explanation:

Domain is the set of x-values for which the function is defined.

Range is the set of y-values for which the function is defined (set of values corresponding to the set of x-values).

The function is given as $y=\frac{1}{2}x+3$

The domain is (-2,0,2,4)

That means, we need to plug-in the 4 numbers given into x of the function and find the 4 corresponding y values (the range).

So, lets do this:

putting x = -2,

$y=\frac{1}{2}(-2)+3=2$

putting x = 0

y = 3

putting x = 2

$y=\frac{1}{2}(2)+3=4$

putting x= 4

$y=\frac{1}{2}(4)+3=5$

Thus, the range is {2,3,4,5}

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Evaluate k to the 6th power times p to the -7 power when k is 2 and p is -1

Solnce55 [7]

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad a^n\implies \cfrac{1}{a^{-n}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
k^6p^{-7}~~
\begin{cases}
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4 years ago

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Answer:

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Explanation:

The median is the middle number. This answer fits best with the question asked.

The mode is the number that occurs the most. This number doesn't really have a correlation to the middle of this data.

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3 years ago

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What two whole square root numbers are between 54

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Step-by-step explanation:

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3 years ago

Anyone knows how to do questions 7 and 8? 15 pts!!

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7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: $2x^2-4x+5$ . With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

$E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3$

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

$2x^2+5=4x\\\\ 2x^2-4x+5=0$

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

$E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)$

Multiplying by (2-1) in the both sides:

$(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ...$ Repeating the process, we obtain: $...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}$

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4 years ago

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