MATHEMATICS HIGH SCHOOL

Column A Column B1. (–22+17)+13 = –22+(17+13)
2. –(27+(–3)) = –27+3
3. 35+(–15+4) = (35+(–15))+4
4. –12+6 = 6+(–12)
a. commutative property
b. associative property
c. opposite of a sum property

Answers

Answer 1
Answer: 1) (-22 + 17) + 13 = -22 + (17+13)
                - 5 + 13 = -22 + 30
                           8 = 8

2) -[27+(-3)] = -27+3
              -24 = -24

3) 35 + (-15+4) = [5+(-15)] + 4
         35 + (-11) = 20 + 4
                     24 = 24

4) -12 + 6 = 6 + (-12)
            - 6 = - 6

Answer: C

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Simplyfy the expression 4x+(15x+y) -11y​

Answers

Steps to solve:

4x + (15x + y) - 11y​

~Remove parenthesis and combine like terms

4x + 15x + y - 11y​

19x - 10y

Best of Luck!

Answer:

(64x -11)/y

Step-by-step explanation:

Multiply 4x into the parenthesis.

60x + 4xy -11y

Divide by y.

(60x + 4x -11)/y

(64x -11)/y

Over a two-hour time period, a snail moved 108 inches. How far is this in yards?

Answers

In two-hour time period, a snail moved 3 yards.

Given that, over a two-hour time period, a snail moved 108 inches.

Metric Conversion refers to the conversion of the given units to desired units for any quantity to be measured.

We know that, 1 yard = 36 inches

Here, in yards = 108/36

= 3 yards

Therefore, in two-hour time period, a snail moved 3 yards.

To learn more about the metric conversion visit:

brainly.com/question/21244256.

#SPJ6
the snail moved 108 inches
108/36=3 yards
as 1 yard=36 in

A container in the shape of a cylinder has a volume of 60 cubic meters. Its base has an area of 15 square units. What is the height of the container?options

3m



2m



4m



5m

Answers

The answer would be 4m

How can you solve -2y+5y=14

Answers

you add the terms that are the same and then simplify.

-2y+5y=3y
3y=14
y=4.666 or 4 2/3

An airplane flew 4 hours with a 25 mph tail wind. The return trip against the same wind took 5 hours. Find the speed of the airplane in still air. This similar to the current problem as you have to consider the 25 mph tailwind and headwind. Plane on outbound trip of 4 hours with 25 mph tailwind and return trip of 5 hours with 25 mph headwind Let r = the rate or speed of the airplane in still air. Let d = the distance a. Write a system of equations for the airplane. One equation will be for the outbound trip with tailwind of 25 mph. The second equation will be for the return trip with headwind of 25 mph.
b. Solve the system of equations for the speed of the airplane in still air.

Answers

Answer:

r  = 225 Mil/h     speed of the airplane in still air

Step-by-step explanation:

Then:

d  is traveled distance   and r  the speed of the airplane in still air

so the first equation is for a 4 hours trip

as  d = v*t

d  =  4 *  ( r  + 25)    (1)          the speed of tail wind  (25 mil/h)

Second equation the trip back in 5 hours

d  =  5  * ( r  - 25 )    (2)

So we got a system of two equation and two unknown variables  d  and

r

We solve it by subtitution

from equation (1)     d  =  4r  + 100

plugging in equation 2

4r  + 100  = 5r -  125    ⇒   -r  =  -225     ⇒    r  = 225 Mil/h

And distance is :

d  =  4*r  +  100         ⇒ d  =  4 * ( 225)  +  100

d  =  900 +   100

d  = 1000 miles

3x²+5x-2=0 by any method

Answers

3x²+5x=2

x(3x+5)=2

x=2  or 3x+5=2
           
            x=3/5  
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