Answer:
21.8
Step-by-step explanation:
1.9 + 9 = 10.9
10.9 x 2 = 21.8
Answer:
The answer is the third equation. A = 250*(1 +0.016)^(0.75)
Step-by-step explanation:
Since Javier deposited $250 into an account with annual interest rate, then as the years passes his account will grow in the manner shown below:
account(0) = 250
account(1) = account(0)*(1 + 1.6/100) = account(0)*(1 + 0.016) = account(0)*1.016
account(2) = account(1)*1.016 = account(0)*1.016*1.016 = account(0)*(1.016)²
account(3) = account(2)*1.016 = account(0)*(1.016)²*1.016 = account(0)*(1.016)³
account(n) = account(0)*(1.016)^n
Where n is the number of years, account(0) is the initial amount. In this case only 9 months have passed, so we need to convert this value to years, dividing it by 12, which is 9/12 = 0.75. The initial amount was 250, so the equation is:
A = 250*(1.016)^(0.75)
The answer is the third equation.
5k/2G is the answer to the problem.
Answer:
The correct answer is option A. Base = 11 cm and height = 13 cm
Step-by-step explanation:
It is given that, height of a triangle is 2 cm more than its base.Then height is increased by 2 cm.Then the area of triangle becomes 82.5 cm²
<u>To find the base an d height of original triangle</u>
Let x be the original base 'b' then the height h = x + 2
New height h = x + 2 + 2 = x + 4
Area = bh/2
82.5= (x(x + 4)/2
165 = x² + 4x
x + 4x - 165 = 0
Solving we get x = 11 and x = -15
Take positive value x = 11
Therefore base = 11 and height = x + 2 = 13 cm
The correct answer is option A. Base = 11 cm and height = 13 cm
Answer:
Step-by-step explanation:
Students use algebra to solve equations (of the form px + q = r and p(x + q) = r where p, q, and r are specific rational numbers); using techniques of making zero (adding the additive inverse) and making one (multiplying by the multiplicative inverse) to solve for the variable.
Students identify and compare the sequence of operations used to find the solution to an equation algebraically, with the sequence of operations used to solve the equation with tape diagrams. They recognize the steps as being the same.
Students solve equations for the value of the variable using inverse operations; by making zero (adding the additive inverse) and making one (multiplying by the multiplicative inverse).
