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3 years ago

What is 6 divided by 0.6

Mathematics

1 answer:

slava [35]3 years ago

7 0

Answer:

Step-by-step explanation:

the answer is 10

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A new sidewalk will be 4 feet wide 200 feet long and filled to a depth of 3 inches (0.25 foot) with concrete. How many cubic yar

solmaris [256]

0.25 ft equals 0.083 yd (0.25/3 ft per yd)
4 ft equals 1.33 yd
200 ft equals 66.6 yd

0.083 x 1.33 x 66.6 = 7.35 cubic yards

5 0

3 years ago

Wellpp, I am kinda lost! The answer I got was 27.

LekaFEV [45]

Answer:

26.7 (1 d.p.)

Step-by-step explanation:

Please see attached picture for full solution.

8 0

3 years ago

Which sentence can represent the inequality One-half x + three-fourths greater-than-or-equal-to 5 and one-fourth?

hodyreva [135]

Answer:

The sum of half of a number and three-fourths is at least five and one quarter

Step-by-step explanation:

It's all about making sense of an English sentence.

"half the sum ..." is different from "the sum of half ..."

And ...

"at least" and "not less than" mean the same (≥)

"not more than" and "at most" mean the same (≤)

_____

Since all of the expressions are written out in words, it is a matter of matching the ideas expressed. That's a problem in reading comprehension, not math.

"One-half x + three-fourths [is] greater-than-or-equal-to 5 and one-fourth"

means the same as ...

"The sum of half of a number and three-fourths is at least five and one quarter"

6 0

4 years ago

Read 2 more answers

Find the expansion of tan x about the point X = 0.

yan [13]

Answer:

$f(x) = x +\frac{1}{3}x^{3}+....$

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that The Maclaurin series is given by:

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

= 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

= 4sec²(x) tan²(x) + 2sec⁴(x)

= 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) = sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

$f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......$

$f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......$

$f(x) = x +\frac{1}{3}x^{3}+....$

Therefore, the expansion of tan x at x = 0 is

$f(x) = x +\frac{1}{3}x^{3}+....$ .

8 0

3 years ago

Simplify 5^4/5^3 5^7 5^-1 1/5 1/5^7

geniusboy [140]

Answer:

5^4-3

5¹

by law

a^m÷a^n= a^m-n

5 0

3 years ago