The correct answer is B.) 4
Answer: The answer that I got was J.
Classic Algebra and its unnecessarily complicated sentence structure. As you may have probably known, Algebra has its own "vocabulary set".
"the length of a rectangle exceeds its width by 6 inches" -> length is 6 in. longer than width -> l= w + 6
Since we're solving for the length and width, let's give them each variables.
length = l = w+6
width = w
The next bit of information is "the area is 40 square inches"
Applying the formula for the area of a rectangle we can set up:
l x w = 40
replace "l", or length, with it's alternate value.
(w+6) x w = 40
distribute
+ 6w = 40
subtract 40 from both sides
+ 6w - 40 = 0
factor
(w - 4)(w + 10) = 0
solve for w
w= 4, or -10
So great, we have 2 values; which one do we choose? Since this problem is referring to lengths and inches, we will have to choose the positive value. There is not such thing as a negative distance in the real world.
We now have half of the problem solved: width. Now we just need to find the length which we can do but substituting it back into the original alternate value of l.
l = w + 6
w=4
l = 4 + 6 = 10
The length is 10 in. and the width is 4 in. Hope this helps!
Answer:
<em>81 </em>
Step-by-step explanation:
25, 36, 49, 64, ...
36 - 25 = 11
49 - 36 = 13 = 11 + 2
64 - 49 = 15 = 13 + 2
64 + 15 + 2 = <em>81</em>
// Input value is usernum.
// This code snippet sums 1 + 3 + 5 + ... + usernum
// The answer is stored in the variable summedvalue.
N = (int) (usernum+1)/2; // maximum number of integers to be summed
int *v = malloc(N*sizeof(int)); // allocate storage for array v
// Calculate the number of loop counts and assign array v..
count = 0;
k = 1;
while (1) {
if (k>usernum) { // do not extend v beyond usernum
break;
}
v(count) = k; // assign an odd integer to v, including usenum
count++;
k += 2; // k is an odd number
if k>usernum { // handle usernum as odd or even
k = usernum;
}
}
n = count; // the size of array v.
// Calculate the sum in a for loop
summedvalue = 0; // initialize summedvalue
for (i=0; i<=n; i++) {
summedvalue += v(i);
}