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2 years ago

14

Draco Malfoy or the Weasley twins?

2 answers:

tekilochka [14]2 years ago

7 0

Answer:

DRACO MALFOY FOR SURE

Step-by-step explanation:

sergij07 [2.7K]2 years ago

6 0

Answer:

DRACO MALFOY

Step-by-step explanation:

THAT'S DADDY RIGHT THERE

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URGENT HELP ME PLEASE

Trava [24]

Answer:

(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(c) $\log_2(\dfrac{64}{8})=3$

(d) $\log_4(\dfrac{64}{16})=1$

(e) $\log_6(36^4)=8$

(f) $\log(100^3)=6$

Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

5 0

3 years ago

1: 63 is 75% of what number?

MissTica

1. It is 84.
x=100%
63=75%

(63*100)/75=84

4 0

3 years ago

Read 2 more answers

What is the area of the shaded region? Use 3.14 for π and round your answer to the nearest tenth.

Musya8 [376]

3.14(10²) - 3.14(6²)
3.14(10² - 6²)
3.14(100 - 36)
3.14(64)
200.96
201

3 0

3 years ago

Read 2 more answers

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

so

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

2 years ago

Which angles are vertical to DEA? Question is located in the attached file (image)

ValentinkaMS [17]

Answer:

∠CEB

Step-by-step explanation:

The vertical angle is one that is directly <em>opposite</em> to the original angle. In this image of two intersecting lines, it looks like the angle ∠CEB is directly across from our angle ∠DEA. So, the angle ∠CEB is <em>vertical to</em> ∠DEA.

Hopefully that was helpful! :)

3 0

2 years ago