Please help someone??

Korvikt [17]

$\frac{x^2+x-30}{x^2-3x+2}\\\\x^2-3x+2=0\\\\x^2-x-2x+2=0\\\\x(x-1)-2(x-1)=0\\\\(x-1)(x-2)=0\iff x-1=0\ \vee\ x-2=0\\\\x=1\ \vee\ x=2\\\\Answer:For\ x=1\ or\ x=2.$

8 0

3 years ago

Consider the functions. F(x)=(x+1)2-4 and g(x)=-4|x+1| which statement compares the range of the functions?

Andreyy89

Answer:

<h2> The fourth</h2>

Step-by-step explanation:

Vertex of f is (-1, -4) so its range is limited to y≥-4

|x+1| is always ≥0 therefore -|x+1| is always ≤0 {4 is insignificant to this - slope doesn't mean in range} so its range is limited to y≤0

3 0

3 years ago

PLSSSSS HELP 15 POINTS

Otrada [13]

Answer: it’s 184

explanation: 184 because the angles across from each other are always the same

8 0

3 years ago

Read 2 more answers

A bag of marbles contains 12 red marbles 8 blue marbles and 5 green marbles. If three marbles are pulled out find each of the pr

worty [1.4K]

Answer: The probability of pulling a red marble is 48%, probability of pulling a blue marble is 32%, probability of pulling a green marble is 20%

and

the probability of pulling three green marbles out with replacement is 0.8%.

Step-by-step explanation: Given that a bag of marbles contains 12 red marbles 8 blue marbles and 5 green marbles.

Let S be the sample space for the experiment of pulling a marble.

Then, n(S) = 12 + 8 + 5 = 25.

Let, E, F and G represents the events of pulling a red marble, a blue marble and a green marble respectively.

The, n(E) = 12, n(F) = 8 and n(G) = 5.

Therefore, the probabilities of each of these three events E, F and G will be

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{12}{25}=\dfrac{12}{25}\times100\%=48\%,\\\\\\P(F)=\dfrac{n(F)}{n(S)}=\dfrac{8}{25}=\dfrac{8}{25}\times100\%=32\%,\\\\\\P(G)=\dfrac{n(G)}{n(S)}=\dfrac{5}{25}=\dfrac{5}{25}\times100\%=20\%.$

Now, the probability of pulling three green marbles out with replacement is given by

$P(G)\times P(G)\times P(G)=\dfrac{5}{25}\times\dfrac{5}{25}\times \dfrac{5}{25}=\dfrac{1}{125}=\dfrac{1}{125}\times100\%=0.8\%.$

Thus, the probability of pulling a red marble is 48%, probability of pulling a blue marble is 32%, probability of pulling a green marble is 20%

and

the probability of pulling three green marbles out with replacement is 0.8%.

4 0

3 years ago

PLEASE HELP WITH GRADE 11 MATH. Make sure you show the formula. Substitute values and show all mathematicall operations! show yo

Degger [83]

3x+y

x

=−3

=−y+3

The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:

\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}

3x+y

3(−y+3)+y

−3y+9+y

−2y

y

=−3

=−12

=6

Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:

\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}

x

=−y+3

=−(6)+3

=−3

The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.

We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.

\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}

3x+y

3(−3)+6

−9+6

−3

=−3

=

?

−3

=

?

−3

=−3

Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:

\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}

7x+10y

−2x+y

=36

=9

In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:

\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}

−2x+y

y

=9

=2x+9

Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:

\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}

7x+10y

7x+10(2x+9)

7x+20x+90

27x+90

3x+10

3x

x

=36

=4

=−6

=−2

Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:

\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}

y

=2x+9

=2(−2)+9

=−4+9

=5

The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.

5 0

2 years ago