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2 years ago

Only answer this if you watch the Mandalorian!

Mathematics

2 answers:

Klio2033 [76]2 years ago

6 0

Answer:

I thought it was great!

Step-by-step explanation:

Your opinions?

matrenka [14]2 years ago

6 0

How you think it was?

You might be interested in

PLEASE HELP ME, FIRST ANSWER WILL GET BRAINLEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Bumek [7]

11. 2x+3x+1=21
5x+1 =21
5x=20
x=4
TU= 2(4) which is 8
UB= 3(4)+1 which is 13

12. 4x-1 +2x-1=5x
6x-2=5x
-2=-1x
2=x

X=2
TU= 7
UB= 3
TB=10

Problem 13:

2x-8=x+17
2x=x+25
x=25
AB =42
BC=42
AC=84

Problem 14:
3x-31
-x+6
=
2x-37 for BC part now we solve for x
x+6=2x-37
6=x-37
43=x
So :
X=43
AB=49
AC=98
BC=49

5 0

2 years ago

Please help me with this !

lakkis [162]

Answer: x intercept : (3/2 , 0)

y intercept: (0, -2)

Step-by-step explanation:

6 0

3 years ago

Multiply / Divide Radical Expressions<br> i need help please.

slava [35]

Answer:

2 with the sqaure root of 12 is = 6.9 simp = 6.9

4 with the sqaure root of 10 is = 12.6 simp= 12.6

sum = 19.5

I hoped this helped! :)

5 0

3 years ago

Find the domain of the function y = 3 tan(23x)

solmaris [256]

Answer:

$\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

In other words, the $x$ in $f(x) = 3\, \tan(23\, x)$ could be any real number as long as $x \ne \displaystyle \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right)$ for all integer $k$ (including negative integers.)

Step-by-step explanation:

The tangent function $y = \tan(x)$ has a real value for real inputs $x$ as long as the input $x \ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

Hence, the domain of the original tangent function is $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

On the other hand, in the function $f(x) = 3\, \tan(23\, x)$ , the input to the tangent function is replaced with $(23\, x)$ .

The transformed tangent function $\tan(23\, x)$ would have a real value as long as its input $(23\, x)$ ensures that $23\, x\ne \displaystyle k\, \pi + \frac{\pi}{2}$ for all integer $k$ .

In other words, $\tan(23\, x)$ would have a real value as long as $x\ne \displaystyle \frac{1}{23} \, \left(k\, \pi + \frac{\pi}{2}\right)$ .

Accordingly, the domain of $f(x) = 3\, \tan(23\, x)$ would be $\mathbb{R} \backslash \displaystyle \left\lbrace \left. \frac{1}{23}\, \left(k\, \pi + \frac{\pi}{2}\right) \; \right| k \in \mathbb{Z} \right\rbrace$ .

4 0

2 years ago

Pre-Calc: Find all the zeros of the function.

uysha [10]

The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)