All categories

3 years ago

Ten less than 3 times a numbers is the same as the number plus 4 please help

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

6 0

7 is the correct answer! :-)

You might be interested in

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

HELP ME PLEASE BE QUICK

Ivenika [448]

Answer: 62.83

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I need help on photo above please!!!

Leya [2.2K]

Option two is the correct answer
Hope this helps :)

4 0

3 years ago

Fit a trigonometric function of the form f(t)=c0+c1sin(t)+c2cos(t)f(t)=c0+c1sin⁡(t)+c2cos⁡(t) to the data points (0,5.5)(0,5.5),

larisa86 [58]

Answer

$f(t)=-0.2+4.1sin(t)+4cos(t)$

Step-By-Step Explanation

Given the function $f(t)=c_0+c_1sin(t)+c_2cos(t)$ .

For each pair (t, f(t)) in the data points (0,5.5), (π/2,0.5), (π,−2.5), (3π/2,−7.5)

$f(0)=c_0+0c_1+c_2=5.5$ .

$f(\pi /2)=c_0+c_1+0c_2=0.5$ .

$f(\pi)=c_0+0sin(t)-c_2=-2.5$ .

$f(3\pi /2)=c_0-c_1+0c_2=-7.5$ .

Expressing this as a system of linear equations in matrix form AX=B

$\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)\left( \begin{array}{c} c_{0} \\ c_{1} \\ c_{2}\\ \end{array} \right)=\left(\begin{array}{c} 5.5 \\ 0.5 \\ -2.5 \\ -7.5 \end{array} \right)$

Where

$A=\left(\begin{array}{ccc} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & -1 \\ 0 & -1 & 0 \end{array} \right)$ ,

$B=\left(\begin{array}{c}5.5\\0.5\\-2.5\\-7.5\end{array} \right)$

$X=\left(\begin{array}{c}c_0\\c_1\\c_2\end{array}\right)$

To determine the values of X, we use the expression

$X=(A^{T}A)^{-1}A^{T}B$

$A^{T}A= \left(\begin{array}{ccc} 3 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right)$

$(A^{T}A)^{-1}= \left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)$

$A^{T}B=\left(\begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$

Therefore:

$X=\left(\begin{array}{ccc} 0.4 & -0.2 & 0 \\ -0.2 & 0.6 & 0 \\ 0 & 0 & 0.5 \end{array} \right)\left( \begin{array}{c} 3.5 \\ 8 \\ 8 \end{array} \right)$