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Jobisdone [24]
3 years ago
10

Do 0.3 and 3.0 have the same value ? explain

Mathematics
2 answers:
Digiron [165]3 years ago
8 0
No, 0.3 is a decimal and 3.0 is a whole number
Eduardwww [97]3 years ago
3 0
Nope they do not have the same value let me show you why.
0.3 as a percent is 30% not that much right? well...
3.0 as a percent is 300% now thats alot.
lets look at these two as fractions
                  30
0.3    =     -------
                 100
               
                 300
3.0    =     -------
                 100
thats 3 times more than 100

and lastly 0.3<3.0 or in other words
0.3 is less than 3.0
hope this helps you:)

merry christmas:D
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It costs a family of two $25 to go to the state fair, and a family of six $65. Find the equation, rate of change, and initial va
motikmotik

Answer:

Step-by-step explanation:

Given:

Family of 2

Cost, C1 = $25

Family of 6

Cost, C2 = $65

Given that 2 points are given above as the following in coordinate form:

A(2,25) and B(6,65)

Rate of change is given as = (y2 - y1)/(x2 - x1)

= (65 - 25)/(6-2)

= 40/4

Therefore,

(y - y1)/(x - x1) = (y2 - y1)/(x2 - x1)

(y - 25)/(65 - 25) = (x - 2)/(6 - 2)

(y - 25)/40 = (x - 2)/4

(y - 25) = 40/4 × (x - 2)

(y - 25) = 10 × (x - 2)

Expanding,

y - 25 = 10x - 20

y = 10x - 20 + 25

y = 10x + 5

6 0
3 years ago
What is the slop of the line represented by the equation y=-2/3-5x​
jek_recluse [69]
The slope is -5x.

Explanation:

Rearrange the equation so it’s in y=mx+b format.

y=-5x-2/3
8 0
3 years ago
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r
Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt

If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

Hence,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

Thus,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

and the Taylor theorem with Lagrange remainder is

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Step-by-step explanation:

5 0
3 years ago
(b)<br> Work out the value of<br> 3 x 10 x 40 000 000<br> Give your answer in standard form,
vfiekz [6]
I think the answer would be, 1200000000
5 0
3 years ago
Read 2 more answers
PLEASE HELP! WILL MARK THE BEST ANSWER BRAINLIEST!
Harlamova29_29 [7]

Answer:B

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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