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3 years ago

Find the domain of the function (f/g)(x) given f (x)=x^2-4x-5 and g(x)=x^2-25

1 answer:

8 0

F + g)(x) = f(x) + g(x); (f - g)(x) = f(x) - g(x): (f · g)(x) = f(x) · g(x) ..., let f(x) = 5x+2 and g(x) = x2-1. 4. f(4)=5(4)+2=22 and g(4)=42-1=15 ... (fog)(x) = f [ g(x) ] = f [ 4x2 ] = sqrt( 4.2 ) = 2 | x |; (gof)(x) = g [f(x) ] = g [ s

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Joe uses 16 pounds of fertilizer in his three gardens. He uses 3/8 of the fertilizer in his first garden and 1/4 of the fertiliz

enot [183]

Answer: The fraction of the fertilizer is left = $\dfrac38.$

Step-by-step explanation:

Given: Fraction of fertilizer used in first garden = $\dfrac38$

Fraction of fertilizer used in second garden = $\dfrac14$

Fraction of fertilizer used in third garden = 1- (Fraction of fertilizer used in first garden)-(Fraction of fertilizer used in second garden)

= $1-\dfrac38-\dfrac14$

$=\dfrac{8-3-2}{8}\\\\=\dfrac{3}{8}$

Hence, the fraction of the fertilizer is left = $\dfrac38.$

7 0

3 years ago

What is the slope of a line that is parallel to the line represented by the equation x – y = 8?

malfutka [58]

Hi there!

Since the slope of a lime that is parallel is always the same, the answer is a. -1.

Since the slope of this equation is -1, the slope of any other parallel line is the same (or else the lines would intersect and not new parallel anymore).

Hope this helps!

7 0

3 years ago

Consider the matrix A. A = 1 0 1 1 0 0 0 0 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of

dusya [7]

Answer with Step-by-step explanation:

We are given that a matrix

$A=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

a.We have to find characteristic polynomial in terms of A

We know that characteristic equation of given matrix $\mid{A-\lambda I}\mid=0$

Where I is identity matrix of the order of given matrix

I= $\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$

Substitute the values then, we get

$\begin{vmatrix}1-\lambda&0&1\\1&-\lambda&0\\0&0&-\lambda\end{vmatrix}=0$

$(1-\lambda)(\lamda^2)-0+0=0$

$\lambda^2-\lambda^3=0$

$\lambda^3-\lambda^2=0$

Hence, characteristic polynomial = $\lambda^3-\lambda^2=0$

b.We have to find the eigen value for given matrix

$\lambda^2(1-\lambda)=0$

Then , we get $\lambda=0,0,1-\lambda=0$

$\lambda=1$

Hence, real eigen values of for the matrix are 0,0 and 1.

c.Eigen space corresponding to eigen value 1 is the null space of matrix $A-I$

$E_1=N(A-I)$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&-1\end{array}\right]$

Apply $R_1\rightarrow R_1+R_3$

$A-I=\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]$

Now,(A-I)x=0[/tex]

Substitute the values then we get

$\left[\begin{array}{ccc}0&0&1\\1&-1&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

Then , we get $x_3=0$

And $x_1-x_2=0$

$x_1=x_2$

Null space N(A-I) consist of vectors

x= $\left[\begin{array}{ccc}x_1\\x_1\\0\end{array}\right]$

For any scalar $x_1$

$x=x_1\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

$E_1=N(A-I)=Span(\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Hence, the basis of eigen vector corresponding to eigen value 1 is given by

$\left[\begin{array}{ccc}1\\1\\0\end{array}\right]$

Eigen space corresponding to 0 eigen value

$N(A-0I)=\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]$

$(A-0I)x=0$

$\left[\begin{array}{ccc}1&0&1\\1&0&0\\0&0&0\end{array}\right]\left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right]=0$

$\left[\begin{array}{ccc}x_1+x_3\\x_1\\0\end{array}\right]=0$

Then, $x_1+x_3=0$

$x_1=0$

Substitute $x_1=0$

Then, we get $x_3=0$

Therefore, the null space consist of vectors

$x=x_2=x_2\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

Therefore, the basis of eigen space corresponding to eigen value 0 is given by

$\left[\begin{array}{ccc}0\\1\\0\end{array}\right]$

5 0

3 years ago

I need to solve this using l'hopital's rule and logarithmic diferentiation.

arlik [135]

Yo sup??

For our convenience let h=x+1

therefore

when x tends to -1, h tends to 0

hence we can rewrite it as

$\lim_{h \to \ 0 } (cos(h))^{(cot(h^2 )}$

This inequality is of the form 1∞

We will now apply the formula

$e^(^g^(^x^)^(^f^(^x^)^-^1^)^)$

plugging in the values of g(x) and f(x)

$e^{lim_{h \to \ 0}{(cot(h)^2(cos(h)-1))}$

express coth² as cosh²/sinh² and also write cosh-1 as 2sin²(h/2)

(by applying the property that cos2x=1-sin²x)

After this multiply the numerator and denominator with h² so that we can apply the property that

$\lim_{x \to \ 0 } sinx/x =1$

Now your equation will look like this.

$e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2)*h^2)/(sin(h)^2*h^2)}$

We will now apply the result

$\lim_{x \to \ 0 } sinx/x =1$

where x=h²

we get

$e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2)}$

we now multiply the numerator and denominator with 4 so that we can say

$\lim_{h^2 \to \ 0 } sin^2(h/2)/(h^2/4) = 1$

$e^{lim_{h \to \ 0}{((cos(h)^2(2sin^2(h/2))/(h^2*4/4)}$

$=e^{lim_{h \to \ 0}{((cos(h)^2*2)/(4)}$

Apply the limits and you will get

$e^{cos(0)^2*2/4$

$=e^{1/2}$

Hope this helps.

7 0

3 years ago

17/18 -3/7= need answer please

Umnica [9.8K]

$\dfrac{17}{18}-\dfrac{3}{7}=\\
\dfrac{119}{126}-\dfrac{54}{126}=\\
\dfrac{65}{126}
$

4 0

3 years ago