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Murljashka [212]
2 years ago
12

If y/5+8=7, what is the value of y/5-2

Mathematics
1 answer:
Andrei [34K]2 years ago
4 0
"<span>y/5+8=7, what is the value of y/5-2?"  Simplify the first equation:  
 
y/5 = 7 - 8, or -1, so y/5 = -1, or y = -5.

Then y/5 - 2 equals -5/5 - 2, or -3.</span>
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Salsk061 [2.6K]
The correct answer is B.) 4

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3 years ago
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vova2212 [387]

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3 years ago
the length of a rectangle exceeds its width by 6 inches and the area is 40 square inches. What are the length and width of the r
Lorico [155]

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

w^{2} + 6w = 40

subtract 40 from both sides

w^{2} + 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!

8 0
3 years ago
Find the next number in the sequence.<br><br>25​, 36​, 49​, 64​, ...
ra1l [238]

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>

6 0
3 years ago
Read 2 more answers
Write a for loop that assigns summedvalue with the sum of all odd values from 1 to usernum. assume usernum is always greater tha
Lapatulllka [165]
// 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);
    }


5 0
3 years ago
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