# Hard math problems solved

## The Best Hard math problems solved

Hard math problems solved can be a useful tool for these scholars. Solving integral equations is a way of finding a function that satisfies a certain equation. In other words, it involves finding a function that "integrates" to a given value. This can be done by using a variety of methods, including integration by parts, integration by substitution, and integration by partial fractions. Each method has its own strengths and weaknesses, and the best method to use will depend on the specific equation that needs to be solved. However, no matter which method is used, solving integral equations can be a challenging task. Fortunately, there are many resources available to help with this process. With a little patience and perseverance, anyone can learn how to solve integral equations.

A parabola is a two-dimensional figure that appears in many mathematical and physical situations. In mathematics, a parabola is defined as a curve where any point is equidistant from a fixed point (called the focus) and a fixed line (called the directrix). In physics, parabolas describe the path of objects under the influence of gravity, such as a ball thrown in the air. In both cases, the equation for a parabola can be quite complicated. However, there are online tools that can help to solve these equations quickly and easily. One such tool is the Parabola Solver, which allows users to input the parameters of their equation and then receive step-by-step instructions for finding the solution. This tool can be an invaluable resource for students and professionals who need to solve complex parabolic equations.

Solving natural log equations requires algebraic skills as well as a strong understanding of exponential growth and decay. The key is to remember that the natural log function is the inverse of the exponential function. This means that if you have an equation that can be written in exponential form, you can solve it by taking the natural log of both sides. For example, suppose you want to solve for x in the equation 3^x = 9. Taking the natural log of both sides gives us: ln(3^x) = ln(9). Since ln(a^b) = b*ln(a), this reduces to x*ln(3) = ln(9). Solving for x, we get x = ln(9)/ln(3), or about 1.62. Natural log equations can be tricky, but with a little practice, you'll be able to solve them like a pro!

Algebra can be used to solve numerous types of problems, including word problems. A word problem is a problem that is expressed in words rather than in mathematical symbols. Many students find that solving word problems is one of the most challenging aspects of college algebra. However, with a little practice, it is possible to master this skill. The key is to read the problem carefully and identify the information that is given and the information that is being asked for. Once this information has been identified, it can be translated into algebraic equations and solved using algebraic methods. With a little practice, solving word problems will become second nature.

How to solve using substitution is best explained with an example. Let's say you have the equation 4x + 2y = 12. To solve this equation using substitution, you would first need to isolate one of the variables. In this case, let's isolate y by subtracting 4x from both sides of the equation. This gives us: y = (1/2)(12 - 4x). Now that we have isolated y, we can substitute it back into the original equation in place of y. This gives us: 4x + 2((1/2)(12 - 4x)) = 12. We can now solve for x by multiplying both sides of the equation by 2 and then simplifying. This gives us: 8x + 12 - 8x = 24, which simplifies to: 12 = 24, and therefore x = 2. Finally, we can substitute x = 2 back into our original equation to solve for y. This gives us: 4(2) + 2y = 12, which simplifies to 8 + 2y = 12 and therefore y = 2. So the solution to the equation 4x + 2y = 12 is x = 2 and y = 2.

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