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3 years ago

50 POINTS !! PLEASE HELP !! ILL GIVE BRAINLIEST TO THE RIGHT ANSWERS.

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

8 0

Answer:

Step-by-step explanation:

you can use the equations for the pythagorean theorem which is a^2 + b^2 =c^2. plug in the numbers you know and then you figure out what yhe missing value is.

You might be interested in

How to solve this? Don't just give me the answer help me learn this.

Sphinxa [80]

Answer:

I'm not entirely sure but I think that what is says is that it wants you to find the value of x in (4x+28) to make it 116 degrees because it is that same angle. So what you do is you get (4x+28)=116 and you do this because they are the same angle so you use 116 and get it to solve x. Since I'm lazy, then I go to this website and type the equation in: m a t h w a y . c o m but without all of the spaces, I had to do that so it would let me answer this, but I'll do it the real way this time. (4x+28)=116 Their is a rule in math that as long as you do something to one side of the equal sign in the equation, you have to do it to the other and it will still work. So, you minus 28 from both sides so that you can get rid of it and you now have 4x=88. Now you should know this part but you divide 4 from both sides to make so x is alone. Now you have x=22 because you divided 4 from each side. So their is your answer: x=22.

Step-by-step explanation: If you have the answer key that you can check from to make sure you got it right because I know some teachers let you do that, then you should make sure that is right, but I am pretty sure it is.

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

An electric current of 0.25 A passes through a circuit that has a resistance of

serious [3.7K]

Answer:

37.5 V

Step-by-step explanation:

A p e x : )

8 0

3 years ago

What are the answers

kolezko [41]

Answer:

I use app called math

way

Step-by-step explanation:

7 0

2 years ago

Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.

Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes = (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0

3 years ago