No, .07 would represent a bigger number if it were coins it would be .06 more pennies. The only time .01 would be greater then .07 is if it were negative
B) square root 62 inches
C) square root 128 inches
x + 2y = 14
x = -26, y = 20
x = 2, y = 6
x = 3, y = 5.5
x = 10, y = 2
Answer:
Option (b) is correct.
x = 2 and y = 6 is the solution to the given system of linear equation.
Step-by-step explanation:
Given: The system of linear equation
2x + 5y = 34
x + 2y = 14
We have to solve the given system of linear equations.
Consider the given system of linear equation
2x + 5y = 34 .....(1)
x + 2y = 14 .......(2)
We solve the syatem using elimination method,
Multiply equation (2) by 2, we get,
(2) ⇒ 2x + 4y = 28 .....(3)
Subtract equation (1) and (3) , we have,
2x + 5y - (2x + 4y) = 34 - 28
Simplify, we get,
y = 6
And put y = 6 in (2), we get,
x + 2(6) = 14
x = 14 - 12
x = 2
Thus, x = 2 and y = 6 is the solution to the given system of linear equation.
x2 – _____
Answer:
The missing value of 64 or 8². The complete expression is x²- 64 or x²- 8².
Step-by-step explanation:
It is given that the difference of squares has a factor of x + 8.
The given expression is
x²- _____
Let the missing term in the given expression be a².
Using the algebraic formula, we get
It is given that one factor of the given expression is (x+8).
It means the value of a is 8 and the missing value of 64 or 8².
Therefore the complete expression is x²- 64 or x²- 8².
Part B: Determine the vertex and indicate whether it is a maximum or a minimum on the graph. How do you know? (2 points)
The function H(t) = −16t2 + 90t + 50 shows the height H(t), in feet, of a projectile after t seconds. A second object moves in the air along a path represented by g(t) = 28 + 48.8t, where g(t) is the height, in feet, of the object from the ground at time t seconds.
Part A: Create a table using integers 1 through 4 for the 2 functions. Between what 2 seconds is the solution to H(t) = g(t) located? How do you know? (6 points)
Part B: Explain what the solution from Part A means in the context of the problem. (4 points)
The correct answers are:
Question 1 - Part A: f(t)=(t+3)²-29; Part B: (-3, -29), minimum; Question 2 - Part A: H(1) = 124, g(1) = 76.8; H(2) = 166, g(2) = 125.6; H(3) = 176, g(3) = 174.4; H(4) = 154, g(4) = 223.2; Part B: Between 3 and 4 seconds, because that is where the values of g(t) catch up with H(t).
Explanation:
Our quadratic function is in the form f(x)=ax²+bx+c. Our value of a is 1, b is 6, and c is -20.
To write a quadratic in vertex form, first take half of the b value and square it: (6/2)² = 3² = 9. This is what we will add and subtract to the function:
f(t) = t²+6t+9-20-9
The squared portion will be (t+b/2)²:
f(t) = (t+3)²-20-9
f(t) = (t+3)²-29
Vertex form is f(x) = a(x-h)²+k, where (h, k) is the vertex; in our function, (h, k) is (-3, -29).
Since the value of a was a positive, this parabola opens upward; this makes the vertex a minimum.
For Question 2 Part A, substitute the values 1, 2, 3 and 4 in H(t) and g(t).
For Part B, we can see that the values of g(t) are much less than that of H(t) until 3 seconds. From there, we can see that g(t) passes H(t). This means that the solution point, where they intersect, is between 3 and 4 seconds.