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3 years ago

Evaluate f(x)=-2x^2-4 for x=3

Mathematics

1 answer:

OLga [1]3 years ago

8 0

I hope this helps you

x=3

f (3)= -2.3^2-4

f (3)= -2.9-4

f (3)= -18-4

f (3)= -22

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Find an equation of the slant asymptote. Do not sketch the curve. y = 5x4 + x2 + x x3 − x2 + 5

murzikaleks [220]

Answer:

The slant asymptote is $y=5 x + 5$ .

Step-by-step explanation:

Line $y=mx+b$ is a slant asymptote of the function $y=f{\left(x \right)}$ , if either $m=\lim_{x \to \infty}\left(\frac{f{\left(x \right)}}{x}\right)=L$ or $m=\lim_{x \to -\infty}\left(\frac{f{\left(x \right)}}{x}\right)=L$ , and L is finite.

We want to find the slant asymptotes of the function

$f(x)=\frac{5 x^{4} + x^{2} + x}{x^{3} - x^{2} + 5}$

First, do polynomial long division

$\frac{5 x^{4} + x^{2} + x}{x^{3} - x^{2} + 5}=5 x + 5+\frac{6 x^{2} - 24 x - 25}{x^{3} - x^{2} + 5}$

Next, we use the above definition,

The first limit is

$\lim_{x \to \infty}\left(\frac{6x^2-24x-25}{x^3-x^2+5}\right)=0$

The second limit is

$\lim_{x \to -\infty}\left(\frac{6x^2-24x-25}{x^3-x^2+5}\right)=0$

The rational term approaches 0 as the variable approaches infinity.

Thus, the slant asymptote is $y=5 x + 5$ .

8 0

3 years ago

Simplify the expression 11x+9-7=?

Vesna [10]

Answer:

11x+2

Step-by-step explanation:

11x+9-7

9-7=2

so 11x+2

b/c u can nat add normal numbers with variable numbers

7 0

4 years ago

Read 2 more answers

I really appreciate the people helping and I was freaking out

yuradex [85]

A Cause I just did that question

3 0

3 years ago

Find the length of side x in simplest radical form with a rational denominator Help please!

myrzilka [38]

Answer:

x = 6

Step-by-step explanation:

x = 3 sin(90) / sin (30)

x = 3(1)/(1/2)

x = 3 ÷ $\frac{1}{2}$

x = 6

8 0

3 years ago

Question -

Sauron [17]

$\rule{200}4$

Answer : The required ratio is (14m-6):(8m+23) .

$\rule{200}4$

Here we are given that the ratio of sum of first n terms of two AP's is (7n + 1):(4n + 27) .

That is.

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{7n +1}{4n +27} \\$

As , we know that the sum of n terms of an AP is given by ,

$\small\sf\longrightarrow \pink{ S_n =\dfrac{n}{2}[2a +(n-1)d]} \\$

Assume that ,

First term of 1st AP = a
First term of 2nd AP = a'
Common difference of 1st AP = d
Common difference of 2nd AP = d'

Using this we have ,

$\small\sf\longrightarrow \dfrac{S_1}{S_2}=\dfrac{\dfrac{n}{2}[2a + (n-1)d]}{\dfrac{n}{2}[2a' +(n-1)d'] } \\$

$\small\sf\longrightarrow \dfrac{7n+1}{4n+27}=\dfrac{2a + (n -1)d}{2a' + (n -1)d' } . . . . . (i) \\$

Now also we know that the nth term of an AP is given by ,

$\longrightarrow\sf\small \pink{ T_n = a + (n-1)d}\\$

Therefore,

$\longrightarrow\sf\small \dfrac{T_{m_1}}{T_{m_2}}= \dfrac{ a + (n-1)d }{a'+(n-1)d'}. . . . . (ii)\\$

$\longrightarrow\sf\small \dfrac{T_1}{T_2}=\dfrac{2a + (2n-2)d}{2a'+(2n-2)d'} . . . . . (iii)\\$

From equation (i) and (iii) ,

$\longrightarrow\sf\small n-1 = 2m-2\\$

$\longrightarrow\sf\small n = 2m -2+1 \\$

$\longrightarrow\sf\small n = 2m -1 \\$

Substitute this value in equation (i) ,

$\longrightarrow \sf\small \dfrac{2a+ (2m-1-1)d}{2a' +(2m-1-1)d'}=\dfrac{7(2m-1)+1}{4(2m-1) +27}\\$

Simplify,

$\longrightarrow\sf\small \dfrac{ 2a + (2m-2)d}{2a' +(2m-2)d'}=\dfrac{14m-7+1}{8m-4+27}\\$

$\longrightarrow\sf\small \dfrac{2[a + (m-1)d]}{2[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

$\longrightarrow\sf\small \dfrac{[a + (m-1)d]}{[a' + (m-1)d']}=\dfrac{ 14m-6}{8m+23}\\$

From equation (ii) ,

$\longrightarrow\sf\small \underline{\underline{\blue{ \dfrac{T_{m_1}}{T_{m_2}}=\dfrac{ 14m-6}{8m+23}}}}\\$

$\rule{200}4$

8 0

3 years ago