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Mariana [72]
3 years ago
11

Can someone please answer this multiple choice question it was only 2 I need it right now I have 10 or 20 minutes left please an

swer it correctly please and please please show work

Mathematics
1 answer:
liq [111]3 years ago
4 0

Answer:

Thanks for reporting me

Step-by-step explanation:

I was just trying to help but ok..... :)

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Find two consecutive odd integers such that 62 more than the lesser is five times the greater.
lidiya [134]
Idk really know srry
8 0
3 years ago
Read 2 more answers
13x+5y=90 what is x? what is y?
Leona [35]

Answer:

X=-5y/13+90/13, Y=-13x/5+18

Step-by-step explanation:

You need to solve the equation for X and Y.

Solving for X:

13x+5y=90

Subtract 5y: 13x=-5y+90

Divide by 13: x=-5y/13+90/13

--

For Y:

13x+5y=90

Subtract 13x: 5y=-13x+90

Divide by 5: y=-13x/5+90/5

(Simplifies to -13x/5+18)

3 0
3 years ago
Answer please I’m dying from math
vovikov84 [41]

Step-by-step explanation:

It's C

I hope it helped:)

8 0
3 years ago
The domain of ​(f​g)(x) consists of the numbers x that are in the domains of both f and g.
Dovator [93]

The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.

Domain is the values of x in the function represented by y=f(x), for which y exists.

THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".

Now we assume the g(x)=x+2 and f(x)=\frac{1}{x-6}

So here since g(x) is a polynomial function so it exists for all real x.

f(x)=\frac{1}{x-6}<em>  </em>does not exists when x=6, so the domain of f(x) is given by all real x except 6.

Now,

(fg)(x)=f(g(x))=f(x+2)=\frac{1}{(x+2)-6}=\frac{1}{x-4}

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.

But domain of both f(x) and g(x) consists of the value x=4.

Hence the statement is not TRUE universarily.

Thus the given statement about the composition of function is FALSE.

Learn more about Domain here -

brainly.com/question/2264373

#SPJ10

3 0
2 years ago
Which equations have infinitely many solutions?
igomit [66]

Answer:

4(x - 1) = 4x - 4

3x + 6 = 3(x + 2)

Step-by-step explanation:

The first equation is

5x - 1 = 5x + 1

We simplify to get;

- 1 = 1

This is not true, therefore this equation has no solution.

The second equation is

3x - 1 = 1 - 3x

Combine like terms:

3x + 3x = 1 + 1

6x = 2

x =  \frac{1}{3}

This has a unique solution.

The 3rd equation is

7x - 2 = 2x - 7

Group similar terms:

7x - 2x =  7 + 2 \\ 5x = 9 \\ x  =  \frac{9}{5}

The 4th equation is :

4(x - 1) = 4x - 4

4x - 4= 4x - 4 \\ x = x

This is always true. The equation has infinite solution.

The 5th equation is:

3x + 6 = 3(x + 2) \\ 3x + 6 = 3x + 6 \\ x = x

This also has infinite solution

The 6th equation is

3(x - 4) = 4(x - 3) \\ 3x - 12 = 4x - 12 \\ 3x - 4x =  - 12 + 12 \\ x = 0

It has a unique solution.

8 0
3 years ago
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