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3 years ago

Im not sure if this is right anyone know the answer ?

Mathematics

1 answer:

Ludmilka [50]3 years ago

3 0

Answer:

you are right

Step-by-step explanation:

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Help!! pls i mark brainly answer

malfutka [58]

Answer:

On the left... True

On the right... False

Step-by-step explanation:

7 0

3 years ago

Please helpp meeee!!!

Law Incorporation [45]

Answer:

im pretty sure the answer would be,

$2 \frac{1}{2}$

Step-by-step explanation:

the median is the middle number of the data, and since there are two medians, you find the mean of the two numbers by adding them together, and then you divide it by 2.

5 0

3 years ago

Find the first, fourth, and tenth terms of the arithmetic sequence described by the given rule.

lubasha [3.4K]

Answer:

The answer is option C.

Step-by-step explanation:

The arithmetic sequence is given by

$A(n) = - 6 + (n - 1)( \frac{1}{5} )$

where n is the number of terms

For the first term

n = 1

So we have

$A(1) = - 6 + (1 - 1)( \frac{1}{5} )$

$= - 6 + (0)( \frac{1}{5} )$

$= - 6$

For the fourth term

n = 4

$A(4) = - 6 + (4 - 1)( \frac{1}{5} )$

$= - 6 + (3)( \frac{1}{5} )$

$= - 5 \frac{2}{5}$

For the tenth term

n = 10

$A(10) = - 6 + (10 - 1)( \frac{1}{5} )$

$= - 6 + (9)( \frac{1}{5} )$

$= - 4 \frac{1}{5}$

Hope this helps you

4 0

3 years ago

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

x = -4 is a vertical asymptote for the function.

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$