What is the solution of the system? Use substitution. -4y-3x=-11 x-2y=17

after 5 points are subtracted from every score, the sample mean is found to be m = 24. what was the mean for the original sampl

VladimirAG [237]

The answer is 29.........................

3 0

3 years ago

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Help with a math problem

Nimfa-mama [501]

Answer:

For B the answer is 1 : 30. For A the graph is Y goes up 2 for every 60 on the X.

3 0

3 years ago

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The line defined by the equation 2y+3=-(2/3)(x-3) is tangent to the graph of g(x) at x=-3. What is the value of the limit as x a

Otrada [13]

the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
2y+3=-(2/3)(x-3)
6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = $\boxed{ -\frac{1}{3} }$

8 0

3 years ago

9/10 ÷ 5/10 can you please explain?

frozen [14]

Answer:

Step-by-step explanation:

Hello,

Dividing by 1/2 is like multiplying par 2, right ?

Dividing by 5/10 is like multiplying par 10/5, right ?

and then

$\dfrac{\dfrac{9}{10}}{\dfrac{5}{10}}=\dfrac{9}{10}\times \dfrac{10}{5}=\dfrac{9}{5}$

as 10/10=1

Thanks

3 0

3 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

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

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

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

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

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

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

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