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3 years ago

Help please I need to know If it is a function of not

Mathematics

1 answer:

algol133 years ago

7 0

Look at the attachment

9. Not a function
10. Is a function
11. Not a function
12. Is a function

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Find the lowest common denominator of p+3/p2+7p+10 and p5/p2+5p+6

Sunny_sXe [5.5K]

(P+1)(p+2)(p+5) is LCM

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3 years ago

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A towns population went from 25800 to 42600 in 15 years. What was the percent of change

Novay_Z [31]

42600-25800=16800
16800/15=1120
42600*x=100*1120
x=112000/42600=2.62%

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4 years ago

<img src="https://tex.z-dn.net/?f=5%5E%7B-3%7D" id="TexFormula1" title="5^{-3}" alt="5^{-3}" align="absmiddle" class="latex-form

AveGali [126]

Answer:

the equivalent equation of

$5^{-3}$ = $\frac{1}{5^{3}}=\frac{1}{125}$

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3 years ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

3 years ago

Help plsplspls asap !! :)

nlexa [21]

Answer:

lets just say the f stands for 1 and 1 times negative 4 is negative 4

and 5 will stay 5 the equation would be -4/5 so after u divide the 4 and the five you get negative 1

8 0

3 years ago

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