All categories

2 years ago

(2n−9)−5(−2.4n+4) Find the difference

Mathematics

1 answer:

goblinko [34]2 years ago

8 0

Answer:

14n - 29

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Combining Like Terms

Step-by-step explanation:

Step 1: Define

(2n - 9) - 5(-2.4n + 4)

Step 2: Simplify

Distribute -5: 2n - 9 + 12n - 20
Combine like terms (n): 14n - 9 - 20
Combine like terms (Z): 14n - 29

You might be interested in

Please help write the equation of the line.

AfilCa [17]

Answer:

x -2y -2=0

Step-by-step explanation:

using the formula, m= ( Y2 - Y1)/( X2-X1)

the gradient is found. the coordinates chosen are (2,0) and ( 0,-1)

the equation of the line is obtained using the formula

y -y1= m(x -x1)

3 0

2 years ago

There are 28 students in Mr. Smith's math class. 6/7 of them prefer pizza over cheeseburgers. How many students prefer pizza?

miv72 [106K]

24 prefer pizza

Plz give brainliest

7 0

3 years ago

Read 2 more answers

Polygon ABCD is translated to create polygon A'B'C'D. Point A and is located at (1, 5) and point A' is located at (-2, 1). What

Ilia_Sergeevich [38]

$\sqrt{ 3^{2}+4^{2} } =5.$
the distance is 5.
hope it helps.

7 0

2 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second