All categories

3 years ago

Kyle crawled 49.8 feet in 6 minutes, how far did he crawl every minute…

Mathematics

1 answer:

Blizzard [7]3 years ago

8 0

Answer:

assuming that kyle crawled at the same pace for six minute then it would be 8.3 feet each minute

Step-by-step explanation:

You might be interested in

2. Barnes and Noble bookstore is having a sale. Dr. Mumin's book sells for

Korvikt [17]

Leannie bought 5 of Dr. Mumin’s book and 3 of Dan Moyer’s book

3 0

3 years ago

To go to the school footsall game, a student would pay $2.00 for each drink purchases,

kotykmax [81]

Answer: 2d

Step-by-step explanation:

You multiply the cost by the amount d, to find the total cost. Since it wants an expression, there is no equal sign.

4 0

3 years ago

Do the equations y = -6x +4 and y = 4x + 6 have one solution, no solution, or infinitely many solutions

Black_prince [1.1K]

Answer:

One solution

Step-by-step explanation:

We know there is only one solution because the lines do not have the same slope, nor are they the exact same line.

So we know they must have just one solution.

Best of luck

6 0

2 years ago

Read 2 more answers

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

3 years ago

Which equation is in standard form?

kifflom [539]

I believe that the last one is true

because the linear equations are

y=ax+b

which means that the point of y-axis is on one side and the slope, point on x-axis and y-intercept are on the other side

good luck

6 0

3 years ago

Read 2 more answers