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2 years ago

The point (0,-4) is located [ANSWER]

Mathematics

2 answers:

tiny-mole [99]2 years ago

8 0

Answer: Its located on the y-axis

Step-by-step explanation: any point where x is zero e.x (0,7) (0,3), is always going to go through the y-axis only

natka813 [3]2 years ago

3 0

Answer:

on the x-axis

Step-by-step explanation:

it is located on the x axis because it is not technically in any quandrant. it is however, on an axis.

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(18 months). hi I'm gothy and I just needed help with this problem. you don't have to include the answer but I'd appreciate it

natulia [17]

Answer:

$93.68

Step-by-step explanation:

Price of computer:

$1486.25

Warranty

$199.99

Total

$1486.25 + $199.99 = $1686.24

Payment per month:

$1686.24/ 18 = $93.68

8 0

2 years ago

Plz help I will give brainliest!

elena55 [62]

Answer:

b, 1 5/8

Step-by-step explanation:

the area is 4 15/32 square meters, and you find the area by doing length x width. the length is provided, 2 3/4. so all you have to do is divide 4 15/32 by 2 3/4 and you get 1 5/8

6 0

2 years ago

Read 2 more answers

Ben and Landon took turns driving on the recent 820 mile road trip Ben averaged 60 mph while Landon average 56 mph if the trip t

AysviL [449]

Answer:

9 hours

Step-by-step explanation:

Let :

Landon hours = x

Ben hours = y

x + y = 14 - - (1)

60x + 56y = 820 - - - (2)

From (1)

y = 14 - x

60x + 56(14 - x) = 820

60x + 784 - 56x = 820

4x = 820 - 784

4x = 36

x = 36/4

x = 9

Landon drove for 9 hours

4 0

2 years ago

Solve x 2 + 8x + 7 = 0 {-1,-7) {1,7) 00

IgorLugansk [536]

X^2 + 8x + 7= 0
(X+7)(X+1)
X=-1,-7

3 0

2 years ago

If f(x) = 9x10 tan−1x, find f '(x).

djverab [1.8K]

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

Step 2: Differentiate

[Function] Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Basic Power Rule: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Arctrig Derivative: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

2 years ago