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2 years ago

Which graph represents the equation y = – (x – 1)2 + 1?

Mathematics

1 answer:

sergey [27]2 years ago

5 0

Answer:

No graph was put in, but heres a helpful answer.

Step-by-step explanation:

There is no graph, but

Y=-1(2x-2)+1

y=-2x+2+1

y=-2x+3

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Sugar expert last year was reduced by 20% from a total of 40000 tonnes. What was the previous years expert?

Readme [11.4K]

Answer:

32000 tonnes

Step-by-step explanation:

Given there was a 20% reduction last year from a total of 40000 tonnes .

20% /100% x 40000

0.2 x 40000

8000tonnes

Therefore, previous year’s excerpt = 40000 - 8000

32000tonnes

6 0

2 years ago

HELP QUICK IM BEING TIMED!!!

DedPeter [7]

Answer:D(38) I think

Step-by-step explanation:

8 0

3 years ago

255.43 divided by 8.9

ruslelena [56]

Answer:

Step-by-step explanation:

The answer is 28.7.

4 0

2 years ago

Read 2 more answers

In order to reflect an exponential function across the

Anna71 [15]

Answer:

take the reciprocal of the initial value "a"

Step-by-step explanation:

From my past learning my teacher have taught me this.

If incorrect let me know

Hope this helps

6 0

3 years ago

Centerville is the headquarters of Greedy Cablevision Inc. The cable company is about to expand service to two nearby towns, Spr

NeX [460]

Answer:

20.2057 Units.

Step-by-step explanation:

First, we determine the length of the cable.

Distance between Centerville (8,0) and point (x,0) is given as:

$\sqrt{(8-x)}^2=8-x$

Distance between point (x,0) and Springfield(0,7) is:

$\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Distance between point (x,0) and Shelvyfield(0,-7) is:

$\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}$

Therefore the Length of the Cable L(x)

$L(x)=(8-x)+2\sqrt{(7)^2+(x)^{2}}$

To find the critical point, we set the derivative of L(x)=0

$L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}$

$\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305$

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

$L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993$

Since $L^{''}(x)$ is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

$L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}$

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

6 0

2 years ago