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andre [41]
3 years ago
13

SOMEONE PLEASE HELP ME ASAP WITH THIS AND SHOW WORK

Mathematics
1 answer:
Phantasy [73]3 years ago
5 0
I’m sorry I can’t help but pray
You might be interested in
5. (9r 3 + 5r 2 + 11r) + (-2r 3 + 9r - 8r 2)
ad-work [718]

Answer:

7r³ - 3r² + 20r

Step-by-step explanation:

(9r³ + 5r² + 11r) + (-2r³ + 9r - 8r²)

You can get rid of the parentheses since there is nothing to distribute.

9r³ + 5r² + 11r - 2r³ + 9r - 8r²

Combine like factors.

7r³ - 3r² + 20r

3 0
3 years ago
HELP ASAP!!!!
solong [7]

Answer:

b

Step-by-step explanation:

6 0
3 years ago
What value of b will cause the system to have an infinite number of solutions?
irga5000 [103]

b must be equal to -6  for infinitely many solutions for system of equations y = 6x + b and -3 x+\frac{1}{2} y=-3

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}

Let us bring the equations in same form for sake of simplicity in comparison

\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}

Now we have two equations  

\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}

Let us first see what is requirement for system of equations have an infinite number of solutions

If  a_{1} x+b_{1} y+c_{1}=0 and a_{2} x+b_{2} y+c_{2}=0 are two equation  

\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} then the given system of equation has no infinitely many solutions.

In our case,

\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}

 As for infinitely many solutions \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}

Hence b must be equal to -6 for infinitely many solutions for system of equations y = 6x + b and  -3 x+\frac{1}{2} y=-3

8 0
3 years ago
Algebra 2
ss7ja [257]
The answer is: g= 6y+8-9/y
4 0
3 years ago
What’s the correct answer for this question?
Jobisdone [24]

Answer:

A.

Step-by-step explanation:

A quadrilateral inscribed in a circle has its opposite angles adding up to 180°

So

<NOP + <M = 180

4x+8x-24 = 180

12x = 180+24

12x = 204

Dividing both sides by 12

x = 17

<NOP = 4(17)

= 68°

8 0
3 years ago
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