Glencoe Algebra 1 - Jenks Public Schools
... RENTALS A hardware store earned $956.50 from renting ladders and power tools last week. The store charged 36 days for ladders and 85 days for power tools. This week the store charged 36 days for ladders, 70 days for power tools, and earned $829. How much does the store charge per day for ladders and ...
... RENTALS A hardware store earned $956.50 from renting ladders and power tools last week. The store charged 36 days for ladders and 85 days for power tools. This week the store charged 36 days for ladders, 70 days for power tools, and earned $829. How much does the store charge per day for ladders and ...
Common Algebra Mistakes
... If the negative is not in parentheses but instead hanging out front of the base, then just bring it down as part of your final answer and proceed to evaluate the exponential expression. The base is negative only if the negative is inside the parentheses and the exponent is outside the parenthese ...
... If the negative is not in parentheses but instead hanging out front of the base, then just bring it down as part of your final answer and proceed to evaluate the exponential expression. The base is negative only if the negative is inside the parentheses and the exponent is outside the parenthese ...
Write and Solve a System of Equations
... Use elimination to solve the system of equations. –3x + 4y = 12 3x – 6y = 18 Since the coefficients of the x-terms, –3 and 3, are additive inverses, you can eliminate the x-terms by adding the equations. Write the equations in column form and add. The x variable is eliminated. Divide each side by –2 ...
... Use elimination to solve the system of equations. –3x + 4y = 12 3x – 6y = 18 Since the coefficients of the x-terms, –3 and 3, are additive inverses, you can eliminate the x-terms by adding the equations. Write the equations in column form and add. The x variable is eliminated. Divide each side by –2 ...
Warmup Quiz - UF Physics
... We solve the first equation for x and plug it in the second equation, and then solve the equation for t: x = 20cos45˚t tan30˚×20cos45˚t +20sin45˚t-4.9t²= t(tan30˚×20cos45˚ +20sin45˚-4.9t) = 0 ⇒ t = 0, (tan30˚×20cos45˚ +20sin45˚)/4.9 = 4.55 Substituting these values in the first equation, we have x = ...
... We solve the first equation for x and plug it in the second equation, and then solve the equation for t: x = 20cos45˚t tan30˚×20cos45˚t +20sin45˚t-4.9t²= t(tan30˚×20cos45˚ +20sin45˚-4.9t) = 0 ⇒ t = 0, (tan30˚×20cos45˚ +20sin45˚)/4.9 = 4.55 Substituting these values in the first equation, we have x = ...
