Answer:
The height of rectangle is 5 inches
Step-by-step explanation:
<u><em>The correct question is</em></u>
A rectangle is drawn so the width is 7 inches longer than the height. If the rectangle’s diagonal measurement is 13 inches, Find the height
Let
x -----> the width of the rectangle in inches
y ----> the height of the rectangle in inches
d ---> diagonal measurement of the rectangle in inches
we know that
Applying the Pythagorean Theorem
we have
so
----> equation A
---> equation B
substitute equation B in equation A
solve for y
solve the quadratic equation by graphing
using a graphing tool
The solution is y=5
see the attached figure
therefore
The height of rectangle is 5 inches
Answer:
y = -2
Step-by-step explanation:
y + 4 = 2
y = 2 - 4
y = -2
Yes just do that.................... ....m
Answer:
Minimum value of function 
is 63 occurs at point (3,6).
Step-by-step explanation:
To minimize :
Subject to constraints:
Eq (1) is in blue in figure attached and region satisfying (1) is on left of blue line
Eq (2) is in green in figure attached and region satisfying (2) is below the green line
Considering 
, corresponding coordinates point to draw line are (0,9) and (9,0).
Eq (3) makes line in orange in figure attached and region satisfying (3) is above the orange line
Feasible region is in triangle ABC with common points A(0,9), B(3,9) and C(3,6)
Now calculate the value of function to be minimized at each of these points.
at A(0,9)
at B(3,9)
at C(3,6)
Minimum value of function is 63 occurs at point C (3,6).
Answer:
D) 0 = 2(x + 5)(x + 3)
Step-by-step explanation:
Which of the following quadratic equations has no solution?
We have to solve the Quadratic equation for all the options in other to get a positive value as a solution for x.
A) 0 = −2(x − 5)2 + 3
0 = -2(x - 5) × 5
0 = (-2x + 10) × 5
0 = -10x + 50
10x = 50
x = 50/10
x = 5
Option A has a solution of 5
B) 0 = −2(x − 5)(x + 3)
Take each of the factors and equate them to zero
-2 = 0
= 0
x - 5 = 0
x = 5
x + 3 = 0
x = -3
Option B has a solution by one of its factors as a positive value of 5
C) 0 = 2(x − 5)2 + 3
0 = 2(x - 5) × 5
0 = (2x -10) × 5
0 = 10x -50
-10x = -50
x = -50/-10
x = 5
Option C has a solution of 5
D) 0 = 2(x + 5)(x + 3)
Take each of the factors and equate to zero
0 = 2
= 0
x + 5 = 0
x = -5
x + 3 = 0
x = -3
For option D, all the values of x are 0, or negative values of -5 and -3.
Therefore the Quadratic Equation for option D has no solution.
