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2 years ago

Is the following statement true or false? A plane has an endpoint.

Mathematics

1 answer:

rjkz [21]2 years ago

8 0

The answer is false because a plane goes on infinitively

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Can anyone help me with this question ?

Assoli18 [71]

I’m pretty sure the answer is k= 10

8 0

3 years ago

PLEASE SOMEONE DO THIS IM RUSHING SO BAD AND IM CRYING - GIVING BRAINLIEST!!!!!

LUCKY_DIMON [66]

Answer:

1.) $1.07

2.) $2.25

3.) $14.20

Step-by-step explanation:

1.) Toonie = 2 CAD

CAD = USD $1.57

$1.57 - $0.50 = $1.07 (USD)

2.) $2.75 - $5 = $2.25 (USD)

3.) $20 x 4 = 80

$80 - $65.80 = $14.20 (USD)

Hope this helped!

8 0

3 years ago

Can someone please help with 4-9

Gnom [1K]

Answer:

make sure you know how to find the square root then divide

Step-by-step explanation:

4 0

2 years ago

A different class has 285 students, and 51.2% of them are men. how many men are in the class?

Zanzabum

Answer:

146

Step-by-step explanation:

Get a percent calculator, Then, you put 51.2% of 285 and you actually get 145.96 but the rounded of what you need is 146

6 0

3 years ago

HELLOOOO HELP PLEASE

MA_775_DIABLO [31]

Answer:

$2*log(x)+log(y)$

Step-by-step explanation:

So, there are two logarithmic identities you're going to need to know.

<em>Logarithm of a power</em>:

$log_ba^c=c*log_ba$

So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: $log_ba=x\implies b^x=a$

Now if we raise both sides to the power of c, we get the following equation: $(b^x)^c=a^c$

Using the exponential identity: $(x^a)^c=x^{a*c}$

We get the equation: $b^{xc}=a^c$

If we convert this back into logarithmic form we get: $log_ba^c=x*c$

Since x was the basic logarithm we started with, we substitute it back in, to get the equation: $log_ba^c=c*log_ba$

Now the second logarithmic property you need to know is

<em>The Logarithm of a Product</em>:

$log_b{ac}=log_ba+log_bc$

Now for a quick proof, let's just say: $x=log_ba\text{ and }y=log_bc$

Now rewriting them both in exponential form, we get the equations:

$b^x=a\\b^y=c$

We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

$b^x*b^y=a*c$

Using the exponential identity: $x^{a}*x^b=x^{a+b}$ , we can rewrite the equation as:

$b^{x+y}=ac$

taking the logarithm of both sides, we get:

$log_bac=x+y$

Since x and y are just the logarithms we started with, we can substitute them back in to get: $log_bac=log_ba+log_bc$

Now let's use these identities to rewrite the equation you gave

$log(x^2y)$

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

$log(x^2)+log(y)$

Now using the logarithm of a power to rewrite the log(x^2) we get:

$2*log(x)+log(y)$

3 0

2 years ago