High-Poverty Charlotte Schools Gets Boost From Duke Energy

High-Poverty Charlotte Schools Gets Boost From Duke Energy Math Tutoring At High-Poverty Charlotte Schools Gets Boost From Duke Energy Math Tutoring At High-Poverty Charlotte Schools Gets Boost From Duke Energy June 3, 2015 Heart is thrilled to share that we are expanding our programs in Charlotte-Mecklenburg Schools, thanks to a $8,000 grant from Duke Energy Foundation. “Math skills that students learn at a young age serve as a foundation block for their entire education career,” said Stick Williams, Duke Energy Foundation president. “Those skills advance other academic pursuits and help our young people meet and overcome new challenges as they prepare for the future.” “Lack of math proficiency can become a barrier to a student’s academic confidence,” said Emily Elliott, executive director of Heart Math Tutoring. “Math deficiencies also affect high school graduation rates and future job opportunities.” The Duke Energy Foundation grant will support program expenses at one of our  school sites, where 50 students, in grades 1-5, receive one-on-one instruction from a volunteer. “We’re very excited that Duke Energy Foundation has chosen to support Heart, and we’re encouraged by the shared focus on K-12 education for underrepresented groups, particularly in the areas of math and science,” Elliott said. “Financial contributions like Duke Energy’s make it possible for us to do this important work.” Read the full press release here.

SAT Math Solving Impossible Math Problems

SAT Math Solving Impossible Math Problems ACT SAT Prep and College Admissions Blog The SAT sometimes approaches math a bit differently than we’re used to in math class. One of these differences in approach is in the way some SAT math problems ask us to solve for relationships between variables, rather than the value of one particular variable. These problems can seem very challenging, or even downright impossible, at first glance, because they often give us multiple variables but only one equation. One thing we learn in math class when studying systems of equations is that in order to solve for 2 variables, we usually need 2 separate equations; to solve for 3 variables, we need 3 separate equations; and so on. Certain SAT math problems will appear to violate that general rule and ask us to do something that’s supposed to be impossible, like solving for 2 variables with only 1 equation. When we look more closely at such problems, though, we realize that the SAT is not really asking for the value of each individual variable involved. Instead, we’re asked to find the value of the sum, product, or some other relationship involving 2 or more variables. Let’s look at an example of such a problem: If 4x + 7y + 1 = y â€" 2, What is the value of 2x + 3y? A. -2 B. -1.5 C. -0.5 D. 2 E. 4.5 Explanation and solution: We’re given only one equation, but it involves 2 unknowns: x and y. We really don’t have enough information to solve for the value of x and/or the value of y, but fortunately, that’s not what the question asks us to do. Instead, we are asked to find the value of a sort of relationship that involves both x and y together: 2x + 3y. To do this, we algebraically manipulate the equation that’s given. Instead of trying to isolate x or isolate y, we try to isolate the quantity “2x + 3y”. To do this, we can first subtract y from both sides of the equation, and then subtract 1 from both sides, so that the equation now looks like: 4x + 6y = -3. To finish isolating “2x + 3y”, we divide both sides of the equation by 2 and obtain: 2x + 3y = -1.5. So the correct answer choice is “B”. Let’s look at another example: If the average of x, y, and z is 12, what is the value of 3(x + y + z)? A. 12 B. 24 C. 36 D. 72 E. 108 Explanation and solution: To obtain the average of x, y, and z, we would need to add the 3 numbers together and then divide the resulting sum by 3. Since we are told that their average equals 12, we can express the information as an equation: (x + y + z)/3 = 12. Multiplying both sides of the equation by 3, we obtain: (x + y + z) = 36. To find the value of 3(x + y + z), we can again multiply both sides of the equation by 3 and find that 3(x + y + z) = 108. The correct answer choice is “E”. Notice that in the problem we just solved, we did not solve for the individual values of x, y, or z; that would have required more information than we were given.