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3 years ago

What equation relates the number of quarters x and the number of quarters x and the number of dime y to the goal of $10

1 answer:

8 0

.25x+.10y=$10

.25 because that is the value of the quarter, times x the number or quarters you have and .10 because that is the value of the dime times y, the number of dimes you have.

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4[(10-4) 2 to the second power divided by 4]

SashulF [63]

<span>4[(10-4) 2 to the second power divided by 4]
=4x6x2</span> to the second power divided by 4
=6x2 to the second power
=6x4
=24
Hope that helps and hope that I didn't misunderstand your question

4 0

3 years ago

Plzzzzz help I don’t understand and this question is worth 23 points that’s how much I need to know how to do this and I need to

butalik [34]

You need PEMDAS
do 3·6 and then 12/6
when you subtract the product and quoetient you get 16 for the answer

6 0

2 years ago

Read 2 more answers

The solution set for the equation x - 6x + 10 = 0 is

Oxana [17]

Answer:

{3-i, 3+ i}

Step-by-step explanation:

hello : the equation is : x²-6x+10 =0

because : (3+i)²-6(3+i)+10 = 9+6i+i²-18-6i+10 =9+6i-1-18-6i+10 =0...right

same method for 3-i... (all quadratic equation have two conjugate solution)

8 0

2 years ago

Four classmates were asked to decorate a bulletin board in the classroom 3/5 of the board was already covered so then divided th

Iteru [2.4K]

Answer:

I think the answer might be 15/100

But im not sure it might be wrong sorry :(

Step-by-step explanation:

5 0

3 years ago

Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$

Hence,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\14z=-14 \end{array}\right.\Rightarrow z=-1$

Substitute it into the second equation:

$23y-9\cdot (-1)=9\Rightarrow 23y+9=9\\ \\23y=0\\ \\y=0$

Substitute them into the first equation:

$3x+2\cdot 0+3\cdot (-1)=3\Rightarrow 3x-3=3\\ \\3x=6\\ \\x=2$

The solution is $x=2,\ y=0,\ z=-1.$

3 0

3 years ago