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stealth61 [152]

3 years ago

6

According to the proportional relationship, how much money did the song bring in from sales in the first week if the pop star ea

rned $800 that week?
Describe what the point (0,0) on the graph represents in terms of the situation being described by the graph.

Which point on the graph represents the amount of money the pop singer gets for $1 in money brought in from sales of the song by the store? (*This is the unit rate)

$800 = x/_____ (For prt.1)

2 answers:

bogdanovich [222]3 years ago

6 0

Answer:

.

$800 = x/1

I think that is correct pelase give me brainlisty

BlackZzzverrR [31]3 years ago

3 0

It’s actually (800,50.00) not 1 I mean I might be wrong but that what I think

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find the slope of the curve y=x^2-2x-5 at the point P(2,5) by finding the limit of secant slopes through point P

Fynjy0 [20]

The point (2, 5) is not on the curve; probably you meant to say (2, -5)?

Consider an arbitrary point Q on the curve to the right of P, $(t,y(t))=(t,t^2-2t-5)$ , where $t>2$ . The slope of the secant line through P and Q is given by the difference quotient,

$\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t$

where we are allowed to simplify because $t\neq2$ .

Then the equation of the secant line is

$y-(-5)=t(x-2)\implies y=t(x-2)-5$

Taking the limit as $t\to2$ , we have

$\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9$

so the slope of the line tangent to the curve at P as slope 2.

- - -

We can verify this with differentiation. Taking the derivative, we get

$\dfrac{\mathrm dy}{\mathrm dx}=2x-2$

and at $x=2$ , we get a slope of $2(2)-2=2$ , as expected.

4 0

3 years ago

Aranibar had 35 in his safety box in January. By thanksgiving, he has $2500. What was his savings rate?

erma4kov [3.2K]

Answer:

71.43%

Step-by-step explanation:

2500/35 = 71.43

Aranibar's savings went up by 71.43%

6 0

3 years ago

Read 2 more answers

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

How many time can 4 go in to 3

vfiekz [6]

Answer:

0.75 times.

Step-by-step explanation:

Hope i helped and can i pls have brainliest.

3 0

3 years ago

Which of the following correctly identifies the set of outputs? (4 points)

spayn [35]

Answer:

the answer is c

Step-by-step explanation:

7 0

3 years ago