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1 year ago

Please help me with this math!!

Mathematics

1 answer:

yan [13]1 year ago

8 0

Answer:

40% increase

Step-by-step explanation:

Initial perimeter:

P = 2(4 + 6) = 20

Perimeter after change:

P = 2(4*2 + 6) = 28

Change in perimeter:

28- 20 = 8

Percent change:

8/20*100% = 40%

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Answer:

Step-by-step explanation:

40 * 5/8 = 200/8 = 25

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2 years ago

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aniked [119]

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2 years ago

I will give brainlest answer quickly please

zvonat [6]

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3 years ago

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Please help i am on a timer

meriva

Answer:

x^2 = 36

Step-by-step explanation:

logx ( 36) = 2

Rewrite this as an exponential equation

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7 0

2 years ago

f p(x) and q(x) are arbitrary polynomials of degreeat most 2, then the mapping< p,q >= p(-2)q(-2)+ p(0)q(0)+ p(2)q(2)defin

Natasha2012 [34]

We're given an inner product defined by

$\langle p,q\rangle=p(-2)q(-2)+p(0)q(0)+p(2)q(2)$

That is, we multiply the values of $p(x)$ and $q(x)$ at $x=-2,0,2$ and add those products together.

$p(x)=2x^2+6x+1$

$q(x)=3x^2-5x-6$

The inner product is

$\langle p,q\rangle=-3\cdot16+1\cdot(-6)+21\cdot(-4)=-138$

To find the norms $\|p\|$ and $\|q\|$ , recall that the dot product of a vector with itself is equal to the square of that vector's norm:

$\langle p,p\rangle=\|p\|^2$

So we have

$\|p\|=\sqrt{\langle p,p\rangle}=\sqrt{(-3)^2+1^2+21^2}=\sqrt{451}$

$\|q\|=\sqrt{\langle q,q\rangle}=\sqrt{16^2+(-6)^2+(-4)^2}=2\sqrt{77}$

Finally, the angle $\theta$ between $p$ and $q$ can be found using the relation

$\langle p,q\rangle=\|p\|\|q\|\cos\theta$

$\implies\cos\theta=\dfrac{-138}{22\sqrt{287}}\implies\theta\approx1.95\,\mathrm{rad}\approx111.73^\circ$

4 0

3 years ago