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Gennadij [26K]
1 year ago
12

Do you know has three times as many dimes as quarters the dollar amount of her dimes and quarters is $6.05 how many quarters doe

s she have
Mathematics
1 answer:
vesna_86 [32]1 year ago
5 0

The number of quarters she have 11.

<h3>What is a system of equations?</h3>

A system of equations is two or more equations that can be solved to get a unique solution. the power of the equation must be in one degree.

Let consider y = quarters; x = dimes

We know that there are 3 times as many dimes as quarters. So we can state that x = 3y.

Then, we say that 25y + 10x = 605

(Value of coin * amount of coins)

Then we substitute x=3y into the equation, yielding:

25y + 10(3y) = 605

25y + 30y = 605

55y = 605

605/55 = 11 = y

Therefore, the number of quarters she have 11.

Learn more about equations here;

brainly.com/question/10413253

#SPJ1

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n200080 [17]

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6 0
3 years ago
A store sells honey in 3/8- quart jars. if the store has 24 quarts if honey available on a shelf, how many jars of honey are on
Kay [80]
24 / (3/8) = 
24 * 8/3 =
192/3 =
64......there are 64 three-eights quart jars
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
find 44% of 8.79. user: find 8% of 2.36. user: darren teaches a class of 25 students. he assigns homework 3 times a week, and ea
Andrews [41]
44% of 8.79 = 
0.44 (8.79) = 3.87 (thats rounded) 

8% of 2.36 =
0.08 (2.36) = 0.19 (rounded)

25 students, homework 3 times per week, 12 problems each assignment 
3(12) = 36 problems per student...36(25) = 900 problems for 25 students.
ur answer is 900

7 0
3 years ago
K here’s another one please help
Luden [163]

Answer:

Both the relations are functions, the correct answer is a.

Step-by-step explanation:

In order to solve this problem we will first find the inverse relation as shown below:

y = 3x^2 + 5\\x = 3y^2 + 5\\3y^2 = x - 5\\y^2 = \frac{x - 5}{3}\\y = \sqrt{\frac{x - 5}{3}} = \frac{\sqrt{x - 5}}{\sqrt{3}}\\y = \frac{\sqrt{x - 5}\sqrt{3}}{\sqrt{3}\sqrt{3}} = \frac{\sqrt{3x - 15}}{3}

Functions are relations between two groups of numbers, for which the input must generate only one output. Using this definition we can classify both the relation and its inverse as a function, therefore the correct answer is a.

7 0
3 years ago
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