Need help with this 4, 5, 6

Jessica has 3 cards in a bag, each marked with a letter. The letters are X, Y, and Z. Without

ahrayia [7]

Answer:

1/27

Step-by-step explanation:

The chance of getting X is ⅓. The chance of getting X then Y is ⅓², or ⅑. Finally, the chance of getting X, then Y, then Z is ⅓³, or 1/27

3 0

2 years ago

Read 2 more answers

Hi could somebody help me

alexandr402 [8]

Assuming both vertical lines are parallel and assuming I'm supposed to find all the unmarked angles...

Angle C:

Supplementary Angles.

x+36=180

180-36=x

x=144

Angle B:

Corresponding Angles.

Angle B ≅ Angle C

144

Angle A:

360 degrees in a quadrilateral

36+144+81+x=360

261+x=360

x = 99

Hope it helps <3

5 0

3 years ago

What is the value of the y-intercept of the graph of h(x) = 1.2(4.15)^x?

gizmo_the_mogwai [7]

9514 1404 393

Answer:

h(0) = 1.2

Step-by-step explanation:

The y-intercept is h(0). When x=0, the exponential term evaluates to 1, so the y-intercept is the multiplier of the exponential term.

h(0) = 1.2 . . . the y-intercept

7 0

2 years ago

Plz help with this math problem

Virty [35]

Re-upload the photo you posted because it isnt showing up :(

4 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.

Logarithm of a power:

$log_ba^c=c*log_ba$

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

The Logarithm of a Product:

$log_b{ac}=log_ba+log_bc$

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