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3 years ago

wite the answer to the problem below, using the correct number of significant figures 18.432+ 5.55 + 19.3

1 answer:

8 0

18.432 + 5.55 + 19.3
We can notice that the last number in the process of addition is 19.3 | with 1 decimal place value / 3 significant figures |
It is better to change all values to 1 decimal place value / 3 significant figures
18.432 → 18.4
5.55 → 5.6 ( 5 rounds up )
19.3 → 19.3
Let us add:
18.4 + 5.6 + 19.3 = 43.3
If you add the original numbers, you will get
18.432 + 5.55 + 19.3 = 43.282
43.282 ≈ 43.3

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Help is highly appreciated

grandymaker [24]

Answer:

Step-by-step explanation:

2-(-1\8)±-7\4 then

2 plus 1\8 plus -7\4

2 plus 1\8 plus -14\8

2 -13\8

2= 16/8

3\8?

3 0

3 years ago

Read 2 more answers

Suppose that a group of 10 students are to be assigned to four dorm rooms: a single, a double, a triple and a quad. How many way

Nesterboy [21]

Answer:

Total number of ways to select all sstudent to make assignment = 10×36×70×1 = 25200

Step-by-step explanation:

We have given number of students = 10

These 10 students are assigned to four dorm rooms a single, a double, a triple and a quad

So number of ways to assigned single student $=^{10}C_1=\frac{10!}{1!9!}=10\\$

Now left student = 10 -1 = 9

So number of ways to assigned double student $=^{9}C_2=\frac{9!}{2!7!}=36$

Now left student = 9-2 = 7

So number of ways to assigned triple student $=^{7}C_3=\frac{7!}{2!4!}=$ =70

Now left student = 7 - 3 = 4

So number of ways to assigned quad student $=^{4}C_4=\frac{4!}{4!0!}=1$

So total number of ways to select all sstudent to make assignment = 10×36×70×1 = 25200

6 0

3 years ago

The sum of three integers is 12

Liula [17]

The three integers are 5,17 and -10

Step-by-step explanation:

Let x, y and z be three integers then

$x+y+z=12 \ \ \Eqn\ 1$

'The sum of the first and second integers exceeds the third by 32'

$x+y=z+32\ \ \ \ Eqn\ 2$

The third integer is 15 less than the first.

$z=x-15\ \ \ \ Eqn\ 3$

From Eqn 2:

$x+y=z + 32\\x+y-z=32$

Subtracting Eqn 1 from eqn 2

$x+y-z-(x+y+z)=32-12\\x+y-z-x-y-z=20\\-2z=20$

Dividing both sides by -2

$\frac{-2z}{-2}=\frac{20}{-2}\\z=-10$

Putting z=-10 in Eqn 3

$z=x-15\\-10=x-15\\x=15-10\\x=5$

Putting x=5 and z=-10 in Eqn 1

$x+y+z=12\\5+y-10=12\\y-5=12\\y=12+5\\y=17\\$

Hence,

The three integers are 5,17 and -10

Keywords: Linear Equations, Variables

Learn more about linear equations at:

#LearnwithBrainly

3 0

3 years ago

Read 2 more answers

What is x + 3/11=51/22

NemiM [27]

$\text{Hey there!}$

$\mathsf{x+\dfrac{3}{11}=\dfrac{51}{22}}\\\\\mathsf{SUBTRACT\ by\ \dfrac{3}{11}\ on\ your\ SIDES}\\\\\mathsf{x+\dfrac{3}{11}-\dfrac{3}{11}=\dfrac{51}{22}-\dfrac{3}{11}}\\\\\mathsf{Cancel\ out: \dfrac{3}{11}-\dfrac{3}{11}\ because\ it\ gives\ you\ the\ result\ of\ 0}\\\\\mathsf{Keep: \dfrac{51}{22}-\dfrac{3}{11}\ because\ it\ gives\ you\ the\ value\ of\ x}\\\\\mathsf{\dfrac{51}{22}-\dfrac{3}{11}=\dfrac{45}{22}}\\\\\boxed{\boxed{\mathsf{Answer:x=\dfrac{45}{22}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\dfrac{\frak{LoveYourselfFirst}}{:)}$

5 0

3 years ago

X-4=4-x

Sidana [21]

Answer:

4 and any number other than 4

Step-by-step explanation:

The first answer is 8. Solve the equation for x.

x-4 = 4-x

To get rid of the x on the right side, you must add x to both sides

(x) + x - 4 = (x) + 4 - x

On the right side x-x=0 so now you are left with 4 on the right side

On the left side x+x=2x so you have 2x-4 on the left side.

Now add 4 to both sides to get rid of the 4 on the left side.

(4) + 2x - 4 = (4) + 4

On the left side, 4-4=0 so you are left with 2x.

On the right side 4+4=8

So now you have 2x = 8. To finish solving for x, you now need to divide both sides by 2.

2x 8

---- = -----

2 2

On the left, 2 divided by 2 equals 1 so you are left with x.

On the right, 8 divided by 2 equals 4.

So now you know what X must equal 4 to make the equation TRUE.

Any number other than 4 will make the equation false.

8 0

3 years ago