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umka21 [38]
3 years ago
5

4 as a product of two primes?

Mathematics
1 answer:
Rama09 [41]3 years ago
8 0

Answer: 2×2=4

Step-by-step explanation: 2 is a prime number so you can do 2×2

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Sean missed 8 questions when he took his quiz for the first time. when he took it the second time he only missed 3 questions. By
Ivanshal [37]

Answer:

37.5

Step-by-step explanation:

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4 years ago
4. Solve the equation for the given variable. For each step. Identity the property used to convert the equation. 22x + 11 = 4x -
Andrei [34K]

Answer:

x=-1

Step-by-step explanation:

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3 years ago
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Given the quadratic function
hjlf

The vertex is (-4,-3)

The axis of symmetry is x=-4

and the transformations are:

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3 years ago
respond to the following in a minimum of 175 of your own words: how would you explain the differences between the trapezoidal ru
harkovskaia [24]

The differences between the trapezoidal rule and simpson's rule is -

The trapezoidal rule and Simpson's method, the latter a set of formulas of varying complexity, are both Newton-Cotes formulas, that are used to examine and model complex curves.

<h3>What is trapezoidal rule?</h3>

The trapezoidal rule is just an integration rule that divides a curve into small trapezoids to calculate the area under it. A area under the curve is calculated by adding the areas of all the small trapezoids.

Follow the steps below to use the trapezoidal rule to determine the area under given curve, y = f. (x).

  • Step 1: Write down the total number of sub-intervals, "n," as well as the intervals "a" and "b."
  • Step 2: Use the formula to determine the width of the sub-interval, h (or) x = (b - a)/n.
  • Step 3: Use the obtained values to calculate this same approximate area of a given curve, ba f(x)dx Tn = (x/2) [f(x0) + 2 f(x1) + 2 f(x2) +....+ 2 f(n-1) + f(n)], where xi = a + ix
<h3>What is Simpson's method?</h3>

Simpson's rule is used to approximate the area beneath the graph of the function f to determine the value of the a definite integral (such that, of the form  b∫ₐ f(x) dx.

Simpson's 1/3 rule provides a more precise approximation. Here are the steps for using Simpson's rule to approximate the integral ba f(x) dx.

  • Step 1: Figure out the values of 'a' & 'b' from interval [a, b], as well as the value of 'n,' which represents the number of subintervals.
  • Step 2: Determine the width of every subinterval using the formula h = (b - a)/n.
  • Step 3: Using the interval width 'h,' divide this same interval [a, b] [x₀, x₁], [x₁, x₂], [x₂, x₃], ..., [xn-2, xn-1], [xn-1, xn] into 'n' subintervals.
  • Step 4: In Simpson's rule formula, substitute all of these values and simplify. b∫ₐ f(x) dx ≈ (h/3) [f(x0)+4 f(x1)+2 f(x2)+ ... +2 f(xn-2)+4 f(xn-1)+f(xn)].

Thus, sometimes we cannot solve an integral using any integration technique, and other times we don't have a particular function to integrate. Simpson's rule aids in approximating the significance of the definite integral in such cases.

To know more about the Simpson's method and trapezoidal rule, here

brainly.com/question/16996659

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3 0
1 year ago
Calculus= Integrate 5^x dx please break down how answer is 1/6x^6+C
aleksley [76]

Answer:

\int\limits {5^x} \, dx = \frac{5^x}{ln\ x} + c

Step-by-step explanation:

Note that the integral of 5^x is not \frac{1}{6}x^6 + c

The solution is as follows:

Given

5^x

Required

Integrate

Represent the given expression using integral notation

\int\limits {5^x} \, dx

This question can't be solved directly;

We'll make use of exponential rules which states;

\int\limits {a^x} \, dx = \frac{a^x}{ln\ x} + c

By comparing \int\limits {5^x} \, dx with \int\limits {a^x} \, dx;

we can substitute 5 for a;

Hence, the expression \int\limits {a^x} \, dx = \frac{a^x}{ln\ x} + c becomes

\int\limits {5^x} \, dx = \frac{5^x}{ln\ x} + c

-------------------------------------------------------------------------------------

However, the integral of x^5 is \frac{1}{6}x^6 + c

This is shown below:

Given that x^5

Applying power rule;

Power rule states that

\int\limits{x^n}\ dx = \frac{x^{n+1}}{n+1} + c

In this case (x^5), n = 5;

So, \int\limits{x^n}\ dx= \frac{x^{n+1}}{n+1} + c

becomes

\int\limits{x^5}\ dx = \frac{x^{5+1}}{5+1} + c

\int\limits{x^5}\ dx = \frac{x^{6}}{6} + c

\int\limits{x^5}\ dx= \frac{x^{6}}{6} + c

4 0
3 years ago
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