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saul85 [17]
3 years ago
6

3) Find two consecutive even integers such that their sum is equal to the difference of 3 times the larger & 2 times the sma

ller.
What do they mean by the larger and the smaller?
Mathematics
1 answer:
swat323 years ago
3 0

Answer:

In assumption, x is the smaller even integer and x+2 is the next so it will be larger.

The integers are 4 and 6

Step-by-step explanation:

Let the two consecutive even integers be x and x+2

Here x is the smaller even integer and x+2 is the next so it will be larger.

Now according to the given statement

Sum of both: x+x+2

difference of 3 times the larger & 2 times the smaller:3(x+2)-2x

Putting the sum and difference equal:

x+x+2 = 3(x+2)-2x

Solving the equation

2x+2 = 3x+6-2x\\2x+2 = x+6\\2x+2-x = x+6-x\\x+2 = 6\\x+2-2 = 6-2\\x = 4

The next integer will be:

x+2 = 4+2 = 6

Hence,

In assumption, x is the smaller even integer and x+2 is the next so it will be larger.

The integers are 4 and 6.

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What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?
MAXImum [283]

Answer:

1

Step-by-step explanation:

6*2-24x+1=

12+1=24x

x+1= 24/12

x+1=2

x=2-1

x=1

4 0
3 years ago
Can someone help me with this
Olin [163]

Answer:

6

Step-by-step explanation:

If p equals 15 then plug 15 in for p and take 15-9=6. pretty sure that's what they are looking for there.

5 0
3 years ago
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What is2/3 divided by 4/5
Sloan [31]
The answer to this question is 5/6
8 0
3 years ago
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Consider the radical equation √c+22 = c + 2.
dimulka [17.4K]

First square both sides to get:

c + 22 = (c + 2)^2

or

c + 22 = c^2 + 4c + 4

Move the terms on the left side to the right side:

c^2 + 3c - 18 = 0

Factor to get:

(c + 6) * (c - 3) = 0.

The solutions are c = -6 and c = 3.

Check to see if these answers work by plugging them into the original equation:

c = -6:

sqrt (-6 + 22) ?= -6 + 2

But, -6 + 2 is a negative number, and you can't get a negative from a square root. So, -6 is extraneous.

c = 3:

sqrt (3 + 22) ?= 3 + 2

5 = 5. So, 3 works.

The answer is: B

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B1-secx%7D%20" id="TexFormula1" title=" \frac{1}{1-secx} " alt=" \frac{1}{1
lidiya [134]
We have the <span> Trigonometric Identities : </span>secx = 1/cosx; (sinx)^2 + (cosx)^2 = 1;
Then, 1 / (1-secx) = 1 / ( 1 - 1/cosx) = 1 / [(cosx - 1)/cosx] = cosx /
(cosx - 1 ) ;
Similar, 1 / (1+secx) = cosx / (1 + cosx) ;
 cosx / (cosx - 1)  +  cosx / (1 + cosx) = [cosx(1 + cosx) + cosx (cosx - 1)] / [ (cosx - 1)(cox + 1)] =[cosx( 1 + cosx + cosx - 1 )] / [ (cosx - 1)(cox + 1)] =  2(cosx)^2 / [(cosx)^2 - (sinx)^2] = <span> 2(cosx)^2 / (-1) = - 2(cosx)^2;
</span>
4 0
3 years ago
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