Who wants points i don't care who

Select the graph of g(x) = |-2х + 6|<br> can someone tell me if this is right

Sauron [17]

Answer:

no

Step-by-step explanation:

the graph should be facing up.

6 0

2 years ago

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What is the expansion of (3+x)^4

Vlad1618 [11]

Answer:

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

Step-by-step explanation:

Considering the expression

$\left(3+x\right)^4$

Lets determine the expansion of the expression

$\left(3+x\right)^4$

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

$a=3,\:\:b=x$

$=\sum _{i=0}^4\binom{4}{i}\cdot \:3^{\left(4-i\right)}x^i$

Expanding summation

$\binom{n}{i}=\frac{n!}{i!\left(n-i\right)!}$

$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

$=\frac{4!}{0!\left(4-0\right)!}\cdot \:3^4x^0+\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1+\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2+\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3+\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4$

as

$\frac{4!}{0!\left(4-0\right)!}\cdot \:\:3^4x^0:\:\:\:\:\:\:81$

$\frac{4!}{1!\left(4-1\right)!}\cdot \:3^3x^1:\quad 108x$

$\frac{4!}{2!\left(4-2\right)!}\cdot \:3^2x^2:\quad 54x^2$

$\frac{4!}{3!\left(4-3\right)!}\cdot \:3^1x^3:\quad 12x^3$

$\frac{4!}{4!\left(4-4\right)!}\cdot \:3^0x^4:\quad x^4$

so equation becomes

$=81+108x+54x^2+12x^3+x^4$

$=x^4+12x^3+54x^2+108x+81$

Therefore,

$\left(3+x\right)^4:\quad x^4+12x^3+54x^2+108x+81$

6 0

3 years ago

The quantity a varies directly with b and c and inversely with dThe quantity d is tripledWhich of the following must be true for

EastWind [94]

Answer:

We have the next relation:

A = (b*d)/c

because we have direct variation with b and d, but inversely variation with c.

Now, if we have 3d instead of d, we have:

A' = (b*(3d))/c

now, we want A' = A. If b,c, and d are the same in both equations, we have that:

3bd/c = b*d/c

this will only be true if b or/and d are equal to 0.

If d remains unchanged, and we can play with the other two variables we have:

3b'd/c' = bd/c

3b'/c' = b/c

from this we can took that: if c' = c, then b' = b/3, and if b = b', then c' = 3c.

Of course, there are other infinitely large possible combinations that are also a solution for this problem where neither b' = b or c' = c

6 0

2 years ago

(2x + 3)" and solve for x

Novosadov [1.4K]

Answer:

2x−3=x

Move all terms containing x

to the left side of the equation.

Subtract x

from both sides of the equation.

2x−3−x=0

Subtract x

from 2x

x−3=0

Add 3

to both sides of the equation.

x=3

7 0

2 years ago

On Friday, the temperature was 82°F. The temperature changed by –2°F on Saturday, and then it changed by 5°F on Sunday. What was

postnew [5]

Answer:

the temp on Sunday was 85, it changed by going down 2 degrees, and up 5 degrees

Step-by-step explanation:

when it says -2 degrees F its basically saying subtract 2, and then go on and add 5

4 0

2 years ago

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