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3 years ago

SHOW WORK PLS I NEED IT ASAP

Mathematics

2 answers:

miss Akunina [59]3 years ago

7 0

Lol I in elementary don’t ask why I’m here

Katena32 [7]3 years ago

6 0

Answer:

i is only in middle school i don't understand but this is a cool question

Step-by-step explanation:

GAEEEEEEEEEEEEEEEEEEEEE!!!!!

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[2 x 2(2 + 1)^2= please help

aleksandrvk [35]

Answer:

36x

Step-by-step explanation:

PEMDAS RULE

3 0

3 years ago

Which of the following functions is not a sinusoid?

Nezavi [6.7K]

A sinusoid is cyclical and has a period or frequency.

y = sin x is a sinusoid, with a period of 2π.
y = √x is not a sinusoid.
y = cos x is a sinusoid with a period of 2π.

Answer: y = √x is not a sinusoid.

8 0

3 years ago

Read 2 more answers

A wedding planner uses 72 ivy stems for 18 centerpieces when she arrives at the venue she realizes she will only need 16 centerp

beks73 [17]

Answer:

64 ivy stems

Step-by-step explanation:

To solve this problem we can use rule of three:

Firstly we had 72 ivy stems to use in 18 centerpieces, and know we have 16 centerpieces and want to know how many ivy stems we need to use, so:

18 centerpieces -> 72 ivy stems

16 centerpieces -> X ivy stems

18/16 = 72/X

X = 16*72/18 = 64

He needs to use now 64 ivy stems to maintain the ratio.

6 0

3 years ago

Read 2 more answers

A. f(x) = l-2xl - 3 Domain: {1, 2, 3} Find the range

gizmo_the_mogwai [7]

Hello there!

We are given the function:

$\displaystyle \large{ f(x) = | - 2x| - 3} \\$

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

x = 1

$\displaystyle \large{ f(1) = | - 2(1)| - 3} \\ \displaystyle \large{ f(1) = | - 2| - 3} \\$

Recall that any numbers in absolute value are always positive.

$\displaystyle \large{ f(1) = 2 - 3} \\ \displaystyle \large{ f(1) = - 1} \\$

x = 2

$\displaystyle \large{ f(2) = | - 2(2)| - 3} \\ \displaystyle \large{ f(2) = | - 4 | - 3} \\ \displaystyle \large{ f(2) = 4 - 3} \\ \displaystyle \large{ f(2) = 1}$

x = 3

$\displaystyle \large{ f(3) = | - 2(3)| - 3} \\ \displaystyle \large{ f(3) = | - 6| - 3} \\ \displaystyle \large{ f(3) = 6 - 3} \\ \displaystyle \large{ f(3) = 3}$

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

6 0

2 years ago

15 Points!

Nezavi [6.7K]

Replace f(x) with 0 and swith the equation around to solve for x

x^4 - 32x^2 - 144 = 0

Factor the left side:

(x+6) (x-6) (x^2+4)

so far we have x+6 = 0, x = -6
x -6 = 0, x = 6

solve x^2 +4 = 0
x^2 = -4
x = sqrt(-4)
x = 2i, -21

The answer is B) 2i, 6, -6

8 0

3 years ago

Read 2 more answers