<h3>Answer:</h3>
room 3 or 4
<h3>Explanation:</h3>
We assume the inequalities tell the rooms that were checked and found empty.
The solution to 3) is ...
... 2x + 3 > 11
... 2x > 8 . . . . . subtract 3
... x > 4 . . . . . . .divide by 2
The solution to 4) is ...
... -3x > -9
... x < 3 . . . . . . divide by -3
Thus, rooms greater than 4 and less than 3 were found empty. Rooms 3 and 4 were not checked, so either one could hold Nessie.
_____
The other inequalities have solutions that are already covered by the solutions to these.
1) x < -1 . . . . after subtacting 4
2) x > 5 . . . . after multiplying by 5
Answer:
1:
p-r/6=-6-(-6)/6=-6+1=<u>-</u><u>5</u>
<u>2</u><u>:</u>
<u>p-</u><u>(</u><u>m-n</u><u>)</u><u>=</u> -1-(-4-4)=-1-(-8)=-1+8=<u>7</u>
3:
(z+y)/2)3=(-5-4)/3=-9/3=<u>-</u><u>3</u>
<u>4</u><u>:</u>
m/6-n=6/6-6=1-6=<u>-</u><u>5</u>
5:
k³-h=3³-(-2)=27+2=<u>2</u><u>9</u>
6:
p-(p-(m-3))=5-(5-(4-3))=5-(5-1)=5-4=<u>1</u>
7:
(k)(k-j)+3=(5)(5-5)+3=5*0+3=<u>3</u>
8.
p²/6-q=6²/6-4=6-4=<u>2</u>
<u>9</u><u>;</u>
zx+3³=6*4+3³=24+27=<u>5</u><u>1</u><u>.</u>
<u>1</u><u>0</u><u>:</u>
<u>y</u><u>+</u><u>z</u><u>-</u><u>(</u><u>y</u><u>-</u><u>x</u><u>)</u><u>=</u>5+4-(5-2)=9-3=<u>6</u>
<u>S</u><u>o</u><u>m</u><u>e</u><u> </u><u>b</u><u>a</u><u>s</u><u>i</u><u>c</u><u> </u><u>r</u><u>u</u><u>l</u><u>e</u><u>s</u><u>:</u>
+. +. +=+
+. * +=+
+. ÷ +=+
+. - +=+
and
+. +. -=add and put sigh of greater one.
-. * +=-sigh
-÷-=+sigh
- - -=add and put - sigh
3x(4x4 – 5x)
= 12x^5-15x^2
Step-by-step explanation:
=== [ -3(x-10)/4] + 2 = 11
==> [( -3x + 30 + 8)/4 = 11
==> -3x = 44 - 38
==> -3x = 6
==> -x = 6/3
==> x = -2
