All categories

3 years ago

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.

You might be interested in

Find the volume of the solid

iogann1982 [59]

Volume = h × l × b

= 3.6 × 2.1 × 1.5

= 11.34 m³

7 0

3 years ago

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:

1/3 (0.33) divide by 3/4 (0.75) = 0.44

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 y is denoted y' )

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

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

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

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

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 e, you can use the change-of-base formula first:

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