Answer:
x ≈ - 1.3
Step-by-step explanation:
16x + 20 = 0
16x + 20 - 20 = 0 - 20
16x = - 20
16x ÷ 16 = - 20 ÷ 16
x = - 1.25

- 1.25 rounds to - 1.3
Answer:
C and D
Step-by-step explanation:
just took the quiz
Answer:
It is divisible by 11 and (a + b) !
Step-by-step explanation:
Given a two digit number
, the digits written in reverse order is
.
Note that a two digit number ab = 10a + b.
For example: 24 = 10(2) + 4
Similarly, ba = 10(b) + a
Now, the sum of the numbers ab and ba = 10a + b + 10b + a
= 11a + 11b
= 11(a + b)
Hence, the sum of any two digit number ab and the reverse of the number ba, is divisible by 11 and (a + b).
Hence, proved.
Answer:
9300
Step-by-step explanation:
To find the total surface are of 150 boxes, all we have to do is multiply the surface area of one box ( 62 which was calculated on this question brainly.com/question/23446865 ) by the total amount of boxes (150)
So your answer = 150 * 62 which equals 9300
To solve this we are going to use the speed equation:

where

is speed

is distance

time
We know from our problem that the upstream trip takes 2 hours, so

. We also know that the downstream trip takes 1.7 hours, so

. Notice that the distance of both trips is the same, so we are going to use

to represent that distance.
Now, lets use our equation to relate the quantities:
For the upstream trip:


equation (1)
For the downstream trip:


equation (2)
We know that the boat travels 2.5 miles per hour faster downstream, so the speed of the boat upstream will be the speed of the boat downstream minus 2.5 miles per hour:

equation (3)
Replacing (3) in (1):


equation (4)
Solving for

in equation (2):


equation (5)
Replacing (5) in (4):







equation (6)
Replacing (6) in (5)



miles
We can conclude that the boat travel

, which is approximately 28.3 miles, in one way.