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timofeeve [1]
4 years ago
8

I just need the answer ​

Mathematics
1 answer:
Alex787 [66]4 years ago
5 0

Answer: y<3

Step-by-step explanation:

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Click on the point that best represents the location of −√5 on the number line.
navik [9.2K]

Answer:

The dot all the way on the left. The one between 2 and 3 I believe.

Step-by-step explanation:

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3 years ago
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The area of a square is 9 centimeters squared (cm^2) If the sides of the square are doubled in length, what is the new area, in
Mice21 [21]

Answer:

The question says area of a square = 9 cm squared.

And they said if the sides are doubled in length what is the new length

so new length = 9 cm^2 × 9 cm^2

which should give you 81cm^2

therefore your answer should be = C

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How many different rays can be formed from five collinear points?
spayn [35]

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3 years ago
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Best Buy is having a HUGE sale. All items are 60% off. How much will you pay for a tv that originally cost $1200 including 5% sa
Whitepunk [10]

Answer:

$480

Step-by-step explanation:

According to the problem, calculation of the given data are as follows,

Original cost including sales tax = $1,200

Discount = 60%

So, we can calculate the cost of TV after discount by using following formula:

Cost of TV after discount =  $1,200 - ( $1,200 × 60%)

= $1,200 - $720

= $480

Hence, Cost of TV after discount is $480.

6 0
3 years ago
Find the absolute extrema of f(x) = e^{x^2+2x}f ( x ) = e x 2 + 2 x on the interval [-2,2][ − 2 , 2 ] first and then use the com
fredd [130]

f(x)=e^{x^2+2x}\implies f'(x)=2(x+1)e^{x^2+2x}

f has critical points where the derivative is 0:

2(x+1)e^{x^2+2x}=0\implies x+1=0\implies x=-1

The second derivative is

f''(x)=2e^{x^2+2x}+4(x+1)^2e^{x^2+2x}=2(2x^2+4x+3)e^{x^2+2x}

and f''(-1)=\frac2e>0, which indicates a local minimum at x=-1 with a value of f(-1)=\frac1e.

At the endpoints of [-2, 2], we have f(-2)=1 and f(2)=e^8, so that f has an absolute minimum of \frac1e and an absolute maximum of e^8 on [-2, 2].

So we have

\dfrac1e\le f(x)\le e^8

\implies\displaystyle\int_{-2}^2\frac{\mathrm dx}e\le\int_{-2}^2f(x)\,\mathrm dx\le\int_{-2}^2e^8\,\mathrm dx

\implies\boxed{\displaystyle\frac4e\le\int_{-2}^2f(x)\,\mathrm dx\le4e^8}

5 0
3 years ago
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