Solve the inequality

1. Consider the system of equations below. What could be the first step in solving by elimination?

leva [86]

Answer:

Multiple the first equation by -1

Step-by-step explanation:

After doing this you add the two equations together and follow the systems of equations as normal

$-3x+-5y=-16\\3x+2y=10\\-3y=-6\\\frac{-3y}{-3} =\frac{-6}{-3} \\y=2$

Therefore y = 2

7 0

3 years ago

-2(x-3) >-8<br> What is the answer

Vlad [161]

Answer:

x < 7

Step-by-step explanation:

First apply the -2 onto each term that it is being multiplied onto.

-2x + 6 > - 8

Now subtract six from both sides.

-2x > -14

Now divide by -2.

x < 7

<u>REMEMBER</u>

When doing greater than and less than equations like this, if you divide by a negative you must flip the greater than/less than.

5 0

3 years ago

19) A six-sided die is rolled six times. What is

aleksley [76]

Answer:

D

Step-by-step explanation:

Probability that the die will show an even number

in one rolling is 3/6=0,5 = p

We have formula for repeating rolling

n!/(r!(n-r)!) p^r (1-p)^(n-r)

n = how many times we repeat rolling = 6

r = how many times we want something to happen

(show an even number exactly two times) = 2

15 *(0,5)^2 *(0,5)^4 = 0,234375

4 0

2 years ago

If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=

Alexxx [7]

$\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\$

$\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}$

now, if we just knew what g(2) is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus 2 = 2x+sin(x)

$\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}$

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it

5 0

2 years ago

The number of lines formed by 4 points, no three of which are collinear, is ____. A.5 B.4 C.6 D.3

djverab [1.8K]

Use the formula
n(n-1)/2

which is the total number of connections between n points

So,
4(3)/2 = 6

the answer is option C.

Hope this helps

6 0

3 years ago