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3 years ago

What is the product α∧4/3·α∧2/3

2 answers:

5 0

$\alpha^\frac{4}{3}\cdot\alpha^\frac{2}{3}\\------------\\use:a^n\cdot a^m=a^{n+m}\\------------\\=\alpha^{\frac{4}{3}+\frac{2}{3}}=\alpha^\frac{4+2}{3}=\alpha^\frac{6}{3}=\boxed{\alpha^2}$

exis [7]3 years ago

4 0

A^⁴/₃ · a^²/₃ = a^(⁴/₃ ⁺ ²/₃) = a^⁶/₃ = a²

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The expression below is the factorization of what trinomal?<br> -1(x+7)(x-4)

Yakvenalex [24]

Fggghfodididididtxitdidotdotd

8 0

3 years ago

The first two terms of an arithmetic sequence are 7 and 4. Find the 7th term.

kobusy [5.1K]

EXPLANATION

If the first two terms of an arithmetic sequence are 7 and 4, then we know that an arithmetic sequence has a constant difference d and is defined by

$a_n=a_1+(n+1)d$

Check wheter the difference is constant:

Compute the differences of all the adjacent terms:

$d=a_{n+1}-a_n$

Replacing terms:

4-7 = -3

The difference between all of the adjacent terms is the same and equal to

d = -3

The first element of the sequence is

$a_1=7$ $a_n=a_1+(n+1)d$

Therefore, the nth term is computed by

d= -3

$a_n=7+\text{ (n-1)}\cdot(-3)$

Refine

d= -3 ,

$a_n=-3n+10$

Now, replacing n=7

$a_7=-3\cdot7+10\text{ = -11}$

So, the answer is -11.

8 0

2 years ago

Select all that apply <br> thαnkѕ pєrѕσn

vodka [1.7K]

Im gonna guess and say the right ones might be
A = AB or A'B'
B = CB or C'B'
D = CA or C'A'
Cus all equal the same side
H0P3 It H2LPS :)
or something

8 0

3 years ago

(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a

Fynjy0 [20]

Step-by-step explanation:

Given that,

X ~ Bernoulli $(p_x)$ and Y ~ Bernoulli $(y_x)$

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

If X and Y can not be both equal to 1

then,

$p_z = P(X=1$ or $Y=1)\\$

$p_z = P(X=1)+P(Y=1)-P(=1$ and $Y =1)$

$p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y$

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0

3 years ago

I need c and d answered badly

Kisachek [45]

C. translated down 2 units

---The "-2" located on the outside of f(x) tells us that the y-values are being changed. With that, the graph can be moved up or down. The presence of the negative/subtraction sign tells us that the graph is moved down.

D. translated right 4 units

---The "-4" located with the x in f(x) tells us that the x-values are being changed. The only tricky thing about this is that the direction of the movement is actually the opposite of the sign. So with the negative/subtraction sign, the graph is moved to the right instead of the left.

Hope this helps!

3 0

3 years ago