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3 years ago

The 180 falls between which two whole numbers? A) 11 and 12 B) 12 and 13 C) 13 and 14 D) 14 and 15

Mathematics

2 answers:

Umnica [9.8K]3 years ago

7 0

Answer:

C) 13 and 14

Step-by-step explanation:

To find the answer, we need to check for all the options.

A) 11 and 12 = 11 *12 = 132

B) 12 and 13 = 12 *13 = 156

C) 13 and 14 = 13*14 = 182

D) 14 and 15 = 14*15 = 210

In which, C) is true.

It is 182 which falls in 180's

Therefore, answer is C) 13 and 14

Hope this will helpful.

Thank you.

alexdok [17]3 years ago

6 0

Answer:

Option C) 13 and 14

Step-by-step explanation:

We are given the following information.

We have to find that 180 lies between the square of two whole numbers.

$a)~ (11)^2\text{ and }(12)^2\\121 \not< 180 \not< 144\\\\b)~ (12)^2\text{ and }(13)^2\\144 \not< 180 \not< 169\\\\c)~ (13)^2\text{ and }(14)^2\\169 < 180 < 196\\\\a)~ (14)^2\text{ and }(15)^2\\196 \not< 180 < 225$

Thus,

180 lies between the square of 13 and 14.

Option C) 13 and 14

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Cloud [144]

Answer:

The value of g(-6) will be "-9".

Step-by-step explanation:

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then,

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On putting the value "-6" at the place of "x", then we get

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3 years ago

Mr. Jones' class counted 500 birds on their field trip last week. The following pie graph shows the different bird species and n

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Answer:

the answers for those are number 4 is 150 and 5 is 16%

Step-by-step explanation:

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15,000 ÷ 100= 150

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3 years ago

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Vitek1552 [10]

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3 years ago

Find the maximum and minimum values of the function f(x,y)=2x2+3y2−4x−5 on the domain x2+y2≤100. The maximum value of f(x,y) is:

ryzh [129]

First find the critical points of f :

$f(x,y)=2x^2+3y^2-4x-5=2(x-1)^2+3y^2-7$

$\dfrac{\partial f}{\partial x}=2(x-1)=0\implies x=1$

$\dfrac{\partial f}{\partial y}=6y=0\implies y=0$

so the point (1, 0) is the only critical point, at which we have

$f(1,0)=-7$

Next check for critical points along the boundary, which can be found by converting to polar coordinates:

$f(x,y)=f(10\cos t,10\sin t)=g(t)=295-40\cos t-100\cos^2t$

Find the critical points of g :

$\dfrac{\mathrm dg}{\mathrm dt}=40\sin t+200\sin t\cos t=40\sin t(1+5\cos t)=0$

$\implies\sin t=0\text{ OR }1+5\cos t=0$

$\implies t=n\pi\text{ OR }t=\cos^{-1}\left(-\dfrac15\right)+2n\pi\text{ OR }t=-\cos^{-1}\left(-\dfrac15\right)+2n\pi$

where n is any integer. We get 4 critical points in the interval [0, 2π) at

$t=0\implies f(10,0)=155$

$t=\cos^{-1}\left(-\dfrac15\right)\implies f(-2,4\sqrt6)=299$

$t=\pi\implies f(-10,0)=235$

$t=2\pi-\cos^{-1}\left(-\dfrac15\right)\implies f(-2,-4\sqrt6)=299$

So f has a minimum of -7 and a maximum of 299.

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3 years ago

I got a question....... LOOK AT ATTACHMENT RIGHT ANSWERS ONLYYYY!!!

enyata [817]

Aw thanks ily too ʕ•ᴥ•ʔ

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2 years ago

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