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Romashka-Z-Leto [24]
3 years ago
10

Find three consecutive even integers such that their sum is 50 more than the largest integer

Mathematics
1 answer:
hram777 [196]3 years ago
4 0
This is how I'd solve it:

1st no. = 2x                (2x)+(2x+2)+(2x+4)=(2x+4)+50
2nd no. = 2x+2                                  6x+6=2x+54
3rd no. = 2x+4                                       4x=48
                                                                x=12

Substitute the x value, and your numbers are <u>24, 26, and 28</u>.  To check: their sum (78) is 50 more than the largest integer (28). 
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Anyone wanna help with this iready question?
Illusion [34]

Answer:

3 ticket = 21 $

5 ticket = 35$

so 1 ticket = 7$

9 ticket = 63$

2$ discount

total amount = 61$

Mark my answer as brainlist answer.

7 0
3 years ago
Whoever can answer all of these will get brainless
inn [45]

Firstly, brainly allows 1 question per question. So do not try this again.

Lets make the fractions all like terms.


242/4 - 37/4=205/4. Lets now make this a mixed fraction.
By doing 205/4, we find out that 4 goes in less than 51 times.

1. 51  1/4.

2. 11/2 = 66/12.....so 66/12 - 29/12= 37/12. Make it mixed fraction: 3 1/12.

3. 36/5=72/10.      72/10 - 37/10 = 35/10. As a mixed it is 3 1/2

4. 35/3 - 17/2 can be rewritten by finding their GCM (Greatest Common Multiple). In this case it is 6.

70/6 - 51/6 = 19/6. 3 1/6.

#LearnWithBrainly, get Verified Answers

5 0
1 year ago
Read 2 more answers
Is this right?......
Kitty [74]
No, the correct answer is square pyramid
4 0
3 years ago
Read 2 more answers
What is the quadratic function that is created with roots at 2 and 4 and a vertex at (3, 1)?
Arada [10]
Hello,

y=k*(x-2)(x-4)
and is passing throught (3,1)
==>1=k*(3-2)(3-4)==>k=-1

y=-(x-2)(x-4) is an answer
4 0
3 years ago
Construct 3 linear equation starting with qiven solution z = 1/3
Andreyy89

Answer:

(a)9z+2=5

(b)21z-11=-4

(c)4z=2-2z

Step-by-step explanation:

We are required to construct 3 linear equations starting with the given solution z = 1/3.

<u>Equation 1</u>

<u />z=\frac{1}{3}<u />

Multiply both sides by 9

9z=\frac{1}{3}\times 9\\9z=3

Rewrite 3 as 5-2

9z=5-2

Add 2 to both sides

Our first equation is: 9z+2=5

<u>Equation 2</u>

<u />z=\frac{1}{3}<u />

Multiply both sides by 21

21z=\frac{1}{3}\times 21\\21z=7

Rewrite 7 as 11-4

21z=11-4

Subtract 11 from both sides

Our second equation is: 21z-11=-4

<u>Equation 3</u>

<u />z=\frac{1}{3}<u />

Multiply both sides by 6

6z=\frac{1}{3}\times 6\\6z=2

Rewrite 6z as 4z+2z

4z+2z=2

Subtract 2z from both sides

Our third equation is: 4z=2-2z

4 0
3 years ago
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