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ale4655 [162]
4 years ago
14

Add or subtract these expressions. Show your work.

Mathematics
1 answer:
fomenos4 years ago
3 0

1. (13x+1) + (2x-6)

(13x + 2x) + (1 + -6)

15x - 5

2. (-12x+8)- (2x-8)

(-12x - 2x) + (8 - -8)

-14x + 16

3.-15+n=-9

-15 + n + 15 = -9 + 15

n = 6

4. v/8 = 2

v/8 * 8 = 2 * 8

v = 16

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ioda

Answer:

2 i think

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Pls do it fast ill mark you brainliest
anygoal [31]

Answer:

\sf  y=7\frac{2}{3} and \sf x = \frac{2}{3}

Explanation:

\sf (1)  \  \  (2x+y = 9)

\sf (1) \ \  (2x+5y = 37)

<u>solve for y</u>:

\sf 2x + y = 9

\sf 2x + 5y = 37

-----------------

\sf 0x + 6y = 46

\sf 6y = 46

\sf  y=\frac{23}{3}

\sf  y=7\frac{2}{3}

<u>solve for x</u>:

\sf 2x + y = 9

\sf 2x = 9 - y

\sf 2x = 9 - \frac{23}{3}

\sf 2x = \frac{4}{3}

\sf x = \frac{2}{3}

8 0
2 years ago
How should this relationship be classified?
Vanyuwa [196]

Answer:

Domain = -2, range = -5

Domain =>0, range = 0

domain = 2, range = 5

let y =>f(x)

It can be seen that Function can be linear having common difference 2

F(x) = 2x + 1 for x>0

is satisfied for above values

let's check

F(2) = 2(2)+1 = 4+1 = 5 As given

For x<0 , F(x) = 2x -1

f(-2)= 2(-2) -1= -4-1= -5 As given

For X= 0, F(x)= 2x

So , F(0) = 0 As given

8 0
3 years ago
Which expression correctly represents “nine less than the quotient of a number and four, increased by three
Natasha2012 [34]

The required expression is \frac{x}{4}-9+3

Step-by-step explanation:

We need to identify the expression correctly represents “nine less than the quotient of a number and four, increased by three"

Solving:

Let the number is x

The quotient of number and four: \frac{x}{4}

nine less than the quotient of a number and four: \frac{x}{4}-9

Increased by 3 : \frac{x}{4}-9+3

So, nine less than the quotient of a number and four, increased by three is:

\frac{x}{4}-9+3

So, The required expression is \frac{x}{4}-9+3

Keywords: Write algebraic expression from words:

Learn more about Write algebraic expression from words at:

  • brainly.com/question/1600376
  • brainly.com/question/1617787
  • brainly.com/question/9720317

#learnwithBrainly

3 0
4 years ago
find all local maximum and minimum points by the second derivative test when possible y=2+3×+3 (b) y=6×+sin3×​
inessss [21]

Answer:

Step-by-step explanation:

Given a function f, whose derivatives are f' and f'', a value x is a critical point if f'(x) =0. A value x is a minimum of f if it is a critical point and f''(x) >0 and it is maximum if f''(x)<0. We will perfom the following steps:

1. Calculate the derivative f'.

2. Solve f'(x) =0.

3. Determine if the x value found in 2 is a minimum or a maximum using f''.

Recall the following properties of derivatives

(\sin(x)) ' = \cos(x)

(\cos(x))' = -\sin(x)

(cf(x))' = cf'(x) where c is a constant.

(x^n)' = nx^{n-1}

(f+g)' = f'+g' where f,g are differentiable.

(c)' =0 where c is a constant.

(f(g(x))' = f'(g(x)) \cdot g'(x) (chain rule)

Case 1: f(x) = 2+3x+3.

Using the properties from above, we have

1. f'(x) = 0+3+0 = 3

2. The equation f'(x)=0 where f'(x) = 3 has no solution.

3. Based on the previous result, f has no maximum nor minimum.

Case 2: f(x) = 6x+\sin(3x)

1. f'(x) = 6+3\cos(3x)

2. We have the equation

6+3\cos(3x)=0

which is equivalent to

\cos(3x) = -2

Recall that the cosine function only takes values in the set [-1,1]. So, this equation has no solution.

3. Based on the previous result, f has no maximum nor minimum.

8 0
3 years ago
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