HELP PLEEEEEEEEEEASE ITS THE LAST QUESTION ON MY TEST ILL DO ANYTHING

What is 3(2x+1)+2(x+3) Help please x

katrin [286]

Answer:

8x + 9

Step-by-step explanation:

Treat the question as 2 separate brackets

First expand 3(2x + 1) :

6x + 3

Then expand 2(x + 3) :

2x + 6

Now add the two answers together and simplify :

= 6x + 3 + 2x + 6

= 8x + 9

hope this helps :)

8 0

2 years ago

Read 2 more answers

A square + b square =

dmitriy555 [2]

c square

its making me write 20 more characters

ignore this

the answer is c squared

8 0

3 years ago

Read 2 more answers

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

2 years ago

Which is the best approximation for <img src="https://tex.z-dn.net/?f=%5Csqrt%7B90%7D" id="TexFormula1" title="\sqrt{90}" alt="\

Aleonysh [2.5K]

Answer:

A9.5

Step-by-step explanation:

I used a calculator......hope it helped:)))))))))))

4 0

2 years ago

Nolan has some nickels and some dimes. He has no more than 19 coins worth no less than $1.30 combined. If Nolan has 5 nickels, d

Allushta [10]

Answer:

number of dimes 11 to 14

Step-by-step explanation:

given data

coin = 19

worth no less = $1.30

Nolan nickles = 5

solution

we convert 5 cent to dollars, it will be

5÷100 = $0.05

and

dime is worth 10 cents

so it will be 10 ÷ 100 = $0.1

so we consider here x as the number when nickels.

and we consider y as the number of dimes.

so here we can say

x + y ≤ 19 ..........................1

as here no less than $1.30 combined.

so

0.05 x + 0.1 y ≥ 1.3 ......................2

and

when Nolan has 5 nickels

put x = 5 in eq 1

5 + y ≤ 19

y ≤ 19 - 5

y ≤ 14

and

now we put x = 5 in eq 2

0.05 × 5 + 0.1y ≥ 1.3

0.25 + 0.1y ≥ 1.3

0.1y ≥ 1.3 - 0.25

0.1y ≥ 1.05

y ≥ 10.5

so all possible values for the number of dimes 11 to 14

7 0

2 years ago