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3 years ago

Simplify. Help me ASAP please

Mathematics

2 answers:

Doss [256]3 years ago

8 0

Answer:

option C is the correct answer my friend

melomori [17]3 years ago

3 0

I’m pretty sure it should be c :)

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Can someone solve 5 + 2x = + 6<br><br> and n - 3n = 14 - 4n<br><br> and 7 ( 5a - 4 ) - 1 = 14 - 8a

professor190 [17]

Answer:

5 + 2x = 6

2x = 6-5

2x = 1

x = 0.5

n -3n = 14 -4n

-2n = 14-4n (since 4n is negative, change the sign to make it positive and then add 4n to both sides of the equation)

2n = 14

n = 7

7 (5a - 4) - 1 = 14 - 8a

35a - 28 - 1 = 14 -8a

35a -29 = 14 - 8a

43a = 43

a = 1

6 0

3 years ago

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Matthew bought 12 roses for his mother. Exactly 1 of the roses were white. How many of the roses were white?

insens350 [35]

<h3>hello!</h3>

What is $\sf{\displaystyle\frac{1}{4}}$ of 12?

In order to find it, we should divide 12 by 4:-

Now, solve our word problem:-

Matthew bought 12 roses. $\sf{\displaystyle\frac{1}{4} }$ of these roses were white.

And we already know how to solve this problem :)

Hence,

$\bigstar{\boxed{\pmb{3~of~the~roses~were~white}}}$

Hope everything is clear; if you need any clarification/explanation, kindly let me know, and I will comment and/or edit my answer :)

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2 years ago

Joe won a lottery jackpot that will pay him $12000.00 eachyear

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Please see attachment

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3 years ago

Given LaTeX: f\left(x\right)=x^{^3}-3x+4f ( x ) = x 3 − 3 x + 4, determine the intervals where the function is increasing and wh

Vedmedyk [2.9K]

Answer:

Increasing: $x$ and $x>1$ .

Decreasing: $-1$

Step-by-step explanation:

We have been given a function $f(x)=x^3-3x+4$ . We are asked to determine the intervals, where the function is increasing and where it is decreasing.

First of all, we will find critical points of our given function by equating derivative of our given function to 0.

Let us find derivative of our given function.

$f'(x)=\frac{d}{dx}(x^3)-\frac{d}{dx}(3x)+\frac{d}{dx}(4)$

$f'(x)=3x^{3-1}-3+0$

$f'(x)=3x^{2}-3$

Let us equate derivative with 0 as find critical points as:

$0=3x^{2}-3$

$3x^{2}=3$

Divide both sides by 3:

$x^{2}=1$

Now we will take square-root of both sides as:

$\sqrt{x^{2}}=\pm\sqrt{1}$

$x=\pm 1$

$x=-1,1$

We know that these critical points will divide number line into three intervals. One from negative infinity to -1, 2nd -1 to 1 and 3rd 1 to positive infinity.

Now we will check one number from each interval. If derivative of the point is greater than 0, then function is increasing, if derivative of the point is less than 0, then function is decreasing.

We will check -2 from our 1st interval.

$f'(-2)=3(-2)^{2}-3=3(4)-3=12-3=9$