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2 years ago

14

For the table, determine whether the relationship is a function. Then represent the relationship using words, an equation, graph

. X Y (0,4) (1,3) (2,2) (3,1)

1 answer:

Flura [38]2 years ago

6 0

Answer:

don't know

Step-by-step explanation:

not really sure

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It is predicted that garden gnome value will go up by 2% each year, forming a geometric progression. Jean-Claude has a garden gn

yawa3891 [41]

Answer:

Step-by-step explanation:

a ) The worth of Jean-Claude's gnome after 1 year

= 80000 x 1.02

= £81600

b ) common ratio of geometric progression

= 81600 / 80000

= 1.02

c ) worth after 10 years

= 80000 x 1.02¹⁰

= 80000 x 1.21899

= £97519.55

d )

120000 = 80000 x $1.02^k$

1.5 = $1.02^k$

ln1.5 = k ln 1.02

k = ln 1.5 / ln 1.02

= .4054 / .0198

= 20.47 years .

8 0

3 years ago

The maximum distance (in nautical miles) that a radio transmitter signal can be sent is represented by the expression 1.23h√

Evgen [1.6K]

Answer:

182.4

this website deleted my paragraph... TLDR; 182.4 is the answer

3 0

2 years ago

Is 1.4 greater than, less than, or equal to -14/7

alekssr [168]

It is greater than -14/7 because first of all, -14/7 is negative and 1.4 positive which means the positive is greater than negative.
also -14/7 is -2
1.4 is greater than -2

7 0

3 years ago

Read 2 more answers

What is the solution for the following system of equations ? <br> 2x + y = 7 <br> x - 2y = 6

-Dominant- [34]

Answer:

(4,-1)

Step-by-step explanation:

3 0

2 years ago

Find the extreme values of f subject to both constraints f(x,y) = 2x^2+3y^2-4x-5, x^2 + y^2 <=16

Softa [21]

$f(x,y)=2x^2+3y^2-4x-5$
$f_x=4x-4=0\implies x=1$
$f_y=6y=0\implies y=0$

$f(x,y)$ has only one critical point at $(x,y)=(1,0)$ . The function has Hessian

$\mathbf H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}4&0\\0&6\end{bmatrix}$

which is positive definite for all $(x,y)$ , which means $f(x,y)$ attains a minimum at the critical point with a value of $f(1,0)=-7$ .

To find the extrema (if any) along the boundary, parameterize it by $x=4\cos t$ and $y=4\sin t$ , with $0\le t$ . On the boundary, we have

$f(x(t),y(t))=F(t)=2(4\cos t)^2+3(4\sin t)^2-4(4\cos t)-5=32\cos^2t+48\sin^2t-16\cos t-5$
$F(t)=35-16\cos t-8\cos2t$

Find the critical points along the boundary:

$F'(t)=16\sin t+16\sin2t=16\sin t+32\sin t\cos t=16\sin t(1+2\cos t)=0$
$\implies t=0,\dfrac{2\pi}3,\pi,\dfrac{4\pi}3$

Respectively, plugging these values into $F(t)$ gives 11, 47, 43, and 47. We omit the first and third, as we can see the absolute extrema occur when $F(t)=47$ .

Now, solve for $x,y$ for both cases:

$t=\dfrac{2\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=2\sqrt3\end{cases}$

$t=\dfrac{4\pi}3\implies\begin{cases}x=4\cos t=-2\\y=4\sin t=-2\sqrt3\end{cases}$

so $f(x,y)$ has two absolute maxima at $(x,y)=(-2,\pm2\sqrt3)$ with the same value of 47.

5 0

2 years ago