Hard Math Problem

The bus then stopped for lunch in a suburb before continuing on a 3 hour tour of countryside at a constant speed of 10mph. The ACT will never expect you to do that, so there must be a better way. As soon as we get to √4, we have something that can be simplified to an integer. Something like √5 gets really messy because there are not two equal fractions that can be multiplied together to equal 5.

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What is the value of the seventh term in the sequence? But because equations of ellipses and circles are fair game for higher-level questions on the ACT, we thought we’d feature one here in our Challenge series!

In a geometric sequence in which all of the terms are positive, the second term is 2 and the fourth term is 10. If you just studied these in school, then it wouldn’t be nearly as tough.

Instead, it’s because they take some of the foundations of math topics you’ve already studied in school, like pre-algebra, algebra, and geometry, and turn them into multi-step processes that may combine concepts from different areas. For the middle part of the trip, we know that D = 10mph x 3 hours, so we know that D = 30 miles. We have to know how low the lowest values in this region are. We know that = – Now, if you happen to have the decimal approximations memorized, you will see that: = – If you don’t have this decimal equivalent memorized, you could think about the triangles in QIV: Even if we can’t directly compare the sizes of the two vertical legs, we definitely can compare the two horizontal legs, and 0.8 is definitely bigger than 0.5.

The shaded region inside the circle and outside the triangle has an area of square centimeters. Now that we have the common ratio, we can move from term to term. Suzanne drove for 2 hours 3 hours so the Total Time was 5 hours. If we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. Since Suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the Average Speed would be closer to 10mph. ANSWER: D To find the average speed of the bus, we know we will need to find the Total Distance and the Total Time, so we can start by using another formula (Distance = Rate x Time) to help us find the pieces we’re missing for each part of the trip. Sine is positive in QII, zero at the negative x-axis, and starts to get more and more negative as we go around through QIII. = – = This does not go down as far as Here’s a diagram with region (D) and the line = – . This leaves (E) by the process of elimination, but let’s verify that this works. The sine is negative one at the bottom, where the unit circle intersects the negative y-axis.

If the students arrived back at Thomas Jefferson High School two hours later, approximately what was the average speed for the entire field trip? The moment you find an irrational number as you count through the series, you can eliminate answer choice A, for example.

Hard Math Problem

Finally, the bus drove 40 miles straight back to the high school. So we just need to count the perfect squares before 50. So that means out of our 50 numbers, 7 can be reduced to integers (or fractions) and 43 are irrational. Bonus hint: If you come across a problem like this on the ACT, and don’t know how to solve it, make sure you eliminate some answer choices! Because every circle has 360 degrees: = Solving this proportion to find angle A gives us x = 32.727272 repeating, or approximately 33°. First of all, we need to remember what rational and irrational numbers are. If sin = – , which of the following could be true about ? Which of the following best describes the graph of y = p(x) in the standard (x,y) coordinate plane? Because a sector is a fraction of a circle, we can use the proportion of the area of the sector to the area of entire circle to find the degree measure of the central angle. What would s have to be so that is divisible by (x 2)? If it is taller than it is wide, it looks like the second one. So let’s get started with the equation for an ellipse: = 1 = 1 If an ellipse is wider than it is tall, our equation looks like the first one. In other words, x = -2 must be a root of the equation. If we want polynomial P(x) to be divisible by (x 2), it must be true that P(– 2) = 0. The function values for p(x) vary directly as x for all real numbers. This information helps us find the fraction of the circle delimited by the triangle. This means that the area of the unshaded sector of the circle inside the triangle must be , since these two regions must add up to the total area of the circle: . So: We are told in the problem that the area of the shaded region is . Knowing that 11 cm is the radius allows us to find the area of the entire circle using the equation .


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