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Lorico [155]
3 years ago
7

Round 64.62306647 to the nearest hundredth ​

Mathematics
1 answer:
DaniilM [7]3 years ago
8 0

Answer:

I think the answer to 64.623

since 3 is in the hundredth place we round to 3.

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Find the volume of the solid
iogann1982 [59]

Volume = h × l × b

= 3.6 × 2.1 × 1.5

= 11.34 m³

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Given that the two triangles are similar, solve for x if AU = 20x + 108, UB = 273, BC = 703, UV = 444, AV = 372 and AC = 589 usi
Ray Of Light [21]
X = 18, correct me if i'm wrong.

4 0
4 years ago
Which question can be answered using the expression 1/3 didvied by 3/4
cupoosta [38]

Answer:

1/3 divided by 3/4 = 0.44

Step-by-step explanation:

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8 0
3 years ago
Can anybody help me solve or give me answers?
laiz [17]
For questions 9 through 12, plug in the term you're looking for in for x:
9. y=10(20)-2=298
10. y=10(9)-2=88
11. y=5(2^5)=160
12.y=5(2^7)=640

For questions 13 through 16, plug the values 1 through 5 in for x, one at a time:
13. y=4^1=4, y=4^2=16, y=4^3=64, y=4^4=256, y=4^5=1,024
14. y=(-3)^1=-3, y=(-3)^2=9, y=(-3)^3=-27, y=(-3)^4=81, y=(-3)^5=-243
15. y=-3^1=-3, y=-3^2=-9, y=-3^3=-27, y=-3^4=-81, y=-3^5=-243
16. y=10^1=10, y=10^2=100, y=10^3=1,000, y=10^4=10,000, y=10^5=100,000

For question 17, plug 4 in for n:
17. f(4)=5^4=625
3 0
3 years ago
Pleeease open the image and hellllp me
Verdich [7]

1. Rewrite the expression in terms of logarithms:

y=x^x=e^{\ln x^x}=e^{x\ln x}

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of <em>y</em> is denoted <em>y' </em>)

y'=e^{x\ln x}(x\ln x)'=x^x(x\ln x)'

y'=x^x(x'\ln x+x(\ln x)')

y'=x^x\left(\ln x+\dfrac xx\right)

y'=x^x(\ln x+1)

2. Chain rule:

y=\ln(\csc(3x))

y'=\dfrac1{\csc(3x)}(\csc(3x))'

y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)

y'=-3\sin(3x)\cot^2(3x)

Since \cot x=\frac{\cos x}{\sin x}, we can cancel one factor of sine:

y'=-3\dfrac{\cos^2(3x)}{\sin(3x)}=-3\cos(3x)\cot(3x)

3. Chain rule:

y=e^{e^{\sin x}}

y'=e^{e^{\sin x}}\left(e^{\sin x}\right)'

y'=e^{e^{\sin x}}e^{\sin x}(\sin x)'

y'=e^{e^{\sin x}+\sin x}\cos x

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to <em>e</em>, you can use the change-of-base formula first:

\log_2x=\dfrac{\ln x}{\ln2}

Then

(\log_2x)'=\left(\dfrac{\ln x}{\ln 2}\right)'=\dfrac1{\ln 2}

So we have

y=\cos^2(\log_2x)

y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'

y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'

y'=-\dfrac2{\ln2}\cos(\log_2x)\sin(\log_2x)

and we can use the double angle identity and logarithm properties to condense this result:

y'=-\dfrac1{\ln2}\sin(2\log_2x)=-\dfrac1{\ln2}\sin(\log_2x^2)

5. Differentiate both sides:

\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'

2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\dfrac{xy'}y+\ln y=0

-\left(2y-\sin x\,e^y-\dfrac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)

y'=\dfrac{2x+\cos x\,e^y\ln y}{2y-\sin x\,e^y-\frac xy}

y'=\dfrac{2xy+\cos x\,ye^y\ln y}{2y^2-\sin x\,ye^y-x}

6. Same as with (5):

\left(\sin(x^2+\tan y)+e^{x^3\sec y}+2x-y+2\right)'=0'

\cos(x^2+\tan y)(x^2+\tan y)'+e^{x^3\sec y}(x^3\sec y)'+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^{x^3\sec y}(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2\right)

y'=-\dfrac{2x\cos(x^2+\tan y)+3x^2\sec y\,e^{x^3\sec y}+2}{\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^{x^3\sec y}-1}

7. Looks like

y=x^2-e^{2x}

Compute the second derivative:

y'=2x-2e^{2x}

y''=2-4e^{2x}

Set this equal to 0 and solve for <em>x</em> :

2-4e^{2x}=0

4e^{2x}=2

e^{2x}=\dfrac12

2x=\ln\dfrac12=-\ln2

x=-\dfrac{\ln2}2

7 0
3 years ago
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