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3 years ago

The equation of a line with a slope of 2 and a y-intercept of -4 is y = -4x + 2.

Mathematics

2 answers:

Bas_tet [7]3 years ago

5 0

Answer: false

Step-by-step explanation:

Nonamiya [84]3 years ago

5 0

False
The slope of the equation is -4 and the y intercept is 2. Use format y = mx + b

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Which property states that you can multiply both sides of an equation by the same term?

Darina [25.2K]

The correct answer is b.
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3 years ago

2. What is the value of the correlation coefficient "r" of the data set?

stiv31 [10]

c) 0.66

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3 years ago

Read 2 more answers

A man is 6 feet 3 inches tall. The tip of his shadow touches a fire hydrant that is 13 feet 6 inches away. What is the angle of

Natalija [7]

well, since a foot is 12inches, 6' 3" will just be 6.25' and 13' 6" will be 13.5'.

Check the picture below.

Make sure your calculator is in Degree mode.

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2 years ago

A candy box is made from a piece of cardboard that measures 15 by 9 inches. Squares of equal size will be cut out of each corner

Valentin [98]

Answer:

1.82 inches

Step-by-step explanation:

Let the size of square piece cut from the corner is d.

So, the length of the box is 15 - 2y

width of the box is 9 - 2y

height of the box is d.

Volume of the box

Volume = length x width x height

V = (15 - 2y)(9 - 2y)y

V = 4y³ - 48y² + 135y

Differentiate with respect to y.

$\frac{dV}{dy}=12y^{2}-96y+135$

It is equal to zero for maxima and minima

$12y^{2}-96y+135=0$

$y=\frac{96\pm \sqrt{96^{2}-4\times 12\times 135}}{24}$

So, y = 1.82 in or 6.2 in

Now

$\frac{d^{2}{V}}{dy^{2}}=24y-96$

For y = 1.82 in, it is negative

and for y = 6.2 in, it is positive

So, the volume of the box is maximum if the height of the box is 1.82 inch.

Thus, the size of the square cut from each of the corner is 1.82 inches.

3 0

4 years ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

3 years ago