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RSB [31]
2 years ago
15

Where did mathematics come from

Mathematics
1 answer:
Ganezh [65]2 years ago
3 0
The most ancient mathematical texts came from mesopotamia and Egypt. The “inventors” of math are the Greek beginning in around the 6th century BC.
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I need help hurry pls
yarga [219]

Answer:

you got this Playa I believe in you

7 0
2 years ago
Read 2 more answers
Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
USPshnik [31]
See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]

M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
     = 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0
3 years ago
Figure out length in inches for brainiest and 5 stars.
otez555 [7]
The answer is 1 and I’m sure about that
3 0
3 years ago
I need help . This problem is confusing me.<br><br>​
Sholpan [36]

Answer:

The answer is 3093.

3093 (if that series you posted actually does stop at 1875; there were no dots after, right?)

Step-by-step explanation:

We have a finite series.

We know the first term is 48.

We know the last term is 1875.

What are the terms in between?

Since the terms of the series form a geometric sequence, all you have to do to get from one term to another is multiply by the common ratio.

The common ratio be found by choosing a term and dividing that term by it's previous term.

So 120/48=5/2 or 2.5.

The first term of the sequence is 48.

The second term of the sequence is 48(2.5)=120.

The third term of the sequence is 48(2.5)(2.5)=300.

The fourth term of the sequence is 48(2.5)(2.5)(2.5)=750.

The fifth term of the sequence is 48(2.5)(2.5)(2.5)(2.5)=1875.

So we are done because 1875 was the last term.

This just becomes a simple addition problem of:

48+120+300+750+1875

168      +   1050  +1875

          1218          +1875

                         3093

6 0
3 years ago
Can someone help me?It's urgent and thank you!
allochka39001 [22]

Answer:

the first option

Step-by-step explanation:

because x^3 is exponential while ab^2 is linear

3 0
3 years ago
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