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Bess [88]
3 years ago
9

What is the sum of two consecutive integers is less than 83. Find the pair of integers with the greatest sum

Mathematics
1 answer:
White raven [17]3 years ago
7 0
Well, the sum has to be less than 83.  Do you think maybe it could be 82 ? ?
Sadly, it couldn't be 82.  If two integers are consecutive, then one is even and
one is odd, and their sum is always odd.  So the greatest sum less than 83
is going to be 81 .

Since the integers are consecutive, both of them have to be right around
half of 81 .  Half of  81  is (40 and 1/2), so let's try  40  and  41 :

<em>40 + 41 = 81  </em>     That's it !       yay!

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Trava [24]
The answer is exponential decay
5 0
3 years ago
Kinda need some help
aksik [14]

Answer:

1) x=10

2) x=6

3) x=13

Step-by-step explanation:

1) 15x-17+4x+7=180

19x-10=180

19x=190

x=10

2) 13x+5=16x-13

5+13=16x-13x

18=3x

x=6

3) 9x+7=11x-19

7+19=11x-9x

26=2x

x=13

6 0
3 years ago
Convert 2845 inches per second to miles per<br> hour. Round to one decimal place.
garik1379 [7]
 the answer would be 161.6 mph
6 0
3 years ago
The measure of angle O is 110°. what is the angle of r and p?
sergij07 [2.7K]

Answer:

Choice D for Angle R

70 degrees for Angle P

Step-by-step explanation:

Google "central angles and inscribed angles geometry" for an explanation.

It's actually a poorly worded question.

7 0
4 years ago
Read 2 more answers
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
3 years ago
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