All categories

2 years ago

240/30 144/3 135/45 108/9

Mathematics

1 answer:

Triss [41]2 years ago

8 0

Answer:

240/30= 8

144/3= 48

135/45= 3

108/9= 12

Step-by-step explanation: these are it simplified aka divided

You might be interested in

HELP PLEASE ASAP!!!<br><br> I NEED HELP AS SOON AS POSSIBLE PLEASE!

Alex

Answer:

c,d,a

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What’s the awnser(can someone do these for me i’m begging you my parents are gonna kill me)

puteri [66]

Answer:

-3/2

(2x-14)/3

(-2,-6)

Step-by-step explanation:

I solved the first one for you already ( -3/2)

To make a line perpindicular to another line they need to intersect at a 90 degree angle. To do this the slope of the lines have to be the negative reciprical (take the fraction, flip it, and apply a negative sign to it)

If you do this to -3/2 you get 2/3

If we follow our y=mx+b equation from before we (so far) have

2/3x+b=y

which means that we just need to solve for b

We also know that it crosses the point (7,0) which means that if we plug in 7 for x and 0 for y we can solve for b

we have

(2/3)*7+b=0

Solve for B

14/3+b=0

b= -14/3

Which means our equation looks like this

(2x-14)/3

our two equations look like this

(-3/2)x-9

(2x-14)/3

We need to find where their coordinates are the same (where they intersect)

We can do this by setting them equal to each other and then solving for x

we have

(-3/2)x-9=(2x-14)/3

solve for x

-4.5x-27=2x-14

-13= 6.5x

x= -2

Now that we know what x equals we can plug this into any of our two beginning equations

(-3/2)*-2-9=y

3-9=y

-6=y

Which means our coordinates are (-2,-6)

if you have any questions let me know

6 0

3 years ago

Hi...Could u plz help me...

DaniilM [7]

Answer:

B and D I dont have time to explain

I am sorry.

Thanks for question

5 0

3 years ago

Differentiate with respect to x and simplify your answer. Show all the appropriate steps? 1.e^-2xlog(ln x)^3 2.e^-2x(log(ln x))^

serious [3.7K]

(1) I assume "log" on its own refers to the base-10 logarithm.

$\left(e^{-2x}\log(\ln x)^3\right)'=\left(e^{-2x}\right)'\log(\ln x)^3+e^{-2x}\left(\log(\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{e^{-2x}}{\ln10(\ln x)^3}\left((\ln x)^3\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10(\ln x)^3}\left(\ln x\right)'$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}(\ln x)^2}{\ln10\,x(\ln x)^3}$

$=-2e^{-2x}\log(\ln x)^3+\dfrac{3e^{-2x}}{\ln10\,x\ln x}$

Note that writing $\log(\ln x)^3=3\log(\ln x)$ is one way to avoid using the power rule.

(2)

$\left(e^{-2x}(\log(\ln x))^3\right)'=(e^{-2x})'(\log(\ln x))^3+e^{-2x}\left(\log(\ln x))^3\right)'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2(\log(\ln x))'$

$=-2e^{-2x}(\log(\ln x))^3+3e^{-2x}(\log(\ln x))^2\dfrac{(\ln x)'}{\ln10\,\ln x}$

$=-2e^{-2x}(\log(\ln x))^3+\dfrac{3e^{-2x}(\log(\ln x))^2}{\ln10\,x\ln x}$

(3)

$\left(\sin(xe^x)^3\right)'=\left(\sin(x^3e^{3x})\right)'=\cos(x^3e^{3x}(x^3e^{3x})'$

$=\cos(x^3e^{3x})((x^3)'e^{3x}+x^3(e^{3x})')$

$=\cos(x^3e^{3x})(3x^2e^{3x}+3x^3e^{3x})$

$=3x^2e^{3x}(1+x)\cos(x^3e^{3x})$

(4)

$\left(\sin^3(xe^x)\right)'=3\sin^2(xe^x)\left(\sin(xe^x)\right)'$

$=3\sin^2(xe^x)\cos(xe^x)(xe^x)'$

$=3\sin^2(xe^x)\cos(xe^x)(x'e^x+x(e^x)')$

$=3\sin^2(xe^x)\cos(xe^x)(e^x+xe^x)$

$=3e^x(1+x)\sin^2(xe^x)\cos(xe^x)$

(5) Use implicit differentiation here.

$(\ln(xy))'=(e^{2y})'$

$\dfrac{(xy)'}{xy}=2e^{2y}y'$

$\dfrac{x'y+xy'}{xy}=2e^{2y}y'$

$y+xy'=2xye^{2y}y'$

$y=(2xye^{2y}-x)y'$

$y'=\dfrac y{2xye^{2y}-x}$

8 0

3 years ago

50 hundred thousands equals how many tenths?

Delicious77 [7]

It would be 100 tens your welcome.

4 0

3 years ago