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3 years ago

Aaron has 13 dimes and 10 quarters. Emily has the same amount of money as Aaron, but she only has nickels and dimes.

Mathematics

2 answers:

Alja [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Aaron has $3.80. Emily has 16 nickels and 30 dimes, which equals $3.80.

guajiro [1.7K]3 years ago

3 0

The awnser is b, 16 nickels, 30 dimes

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Tomas is making trail mix using granola and walnuts. He can spend a total of $12 on the ingredients. He buys 3 pounds of granola

Artist 52 [7]

Answer:

Tomas added 6 to both sides of the equation instead of subtracting 6.

Step-by-step explanation:

Tomas is making trail mix using granola and walnuts. He can spend a total of $12 on the ingredients. He buys 3 pounds of granola that costs $2.00 per pound. The walnuts cost $6 per pound. He uses the equation 2x + 6y = 12 to represent the total cost, where x represents the number of pounds of granola and y represents the number of pounds of walnuts. He solves the equation for y, the number of pounds of walnuts he can buy.

Given:

2x + 6y = 12

where

x = number of pounds of granola y = number of pounds of walnuts

The correct solution to the problem

x = 3 pounds

2x + 6y = 12

2(3) + 6y = 12

6 + 6y = 12

Subtract 6 from both sides

6 + 6y - 6 = 12 - 6

6y = 6

Divide both sides by 6

y = 6/6

= 1

y = 1 pound

Tomas added 6 to both sides of the equation instead of subtracting 6.

3 0

4 years ago

Read 2 more answers

What is 1/6 + 1/6 + 1/6

tatyana61 [14]

Answer:

1/6 + 1/6 + 1/6 =

1/2 or 0.5

6 0

4 years ago

Read 2 more answers

Find the absolute maximum and minimum values of f on the set D. f(x, y) = x3 − 3x − y3 + 12y + 1, D is quadrilateral whose verti

mojhsa [17]

$D$ is the set of points,

$\left\{(x,y)\mid-2\le x\le2,x\le y\le3\right\}$

Check for critical points:

$f(x,y)=x^3-3x-y^3+12y+1\implies\begin{cases}f_x=3x^2-3=0\implies x=\pm1\\f_y=-3y^2+12=0\implies y=\pm2\end{cases}$

Of these 4 points, only 2 belong to $D$ , (-1, 2) and (1, 2), for which we have

$\begin{cases}f(-1,2)=19\\f(1,2)=15\end{cases}$

Now look for extrema along the boundary.

If $x=-2$ , then

$f(-2,y)=-y^3+12y-1\implies f'(-2,y)=-3y^2+12=0\implies y\pm2$

We have $f'(-2,y)>0$ for $-2$ and $f'(-2,y)$ for $2$ , which indicates a local maximum at $y=2$ and minima at the endpoints of this boundary. So

$\begin{cases}f(-2,2)=15\\f(-2,-2)=-17\\f(-2,3)=8\end{cases}$

If $x=2$ , then

$f(2,y)=y^3+12y+3\implies f'(2,y)=-3y^2+12=0\implies y=\pm2$

We have $f'(2,y)$ for $2$ , so we have extrema at the endpoints of this boundary.

$\begin{cases}f(2,2)=19\\f(2,3)=12\end{cases}$

If $y=x$ , then

$f(x,x)=9x+1\implies f'(x,x)=9>0$

which tells us $f$ is strictly increasing on this boundary, giving the extrema we already know about,

$\begin{cases}f(-2,-2)=-17\\f(2,2)=19\end{cases}$

If $y=3$ , then

$f(x,3)=x^3-3x+10\implies f'(x,3)=3x^2-3=0\implies x=\pm1$

We have $f'(x,3)>0$ for $-2$ and $1$ , and $f'(x,3)$ for $-1$ . This indicates a maximum at $x=-1$ and a minimum at $x=1$ , with

$\begin{cases}f(-2,3)=8\\f(-1,3)=12\\f(1,3)=8\\f(2,3)=12\end{cases}$

From this analysis, we find that $f$ attains an absolute maximum of 19 at (-1, 2) and (2, 2), and an absolute minimum of -17 at (-2, -2).

4 0

3 years ago

A linear equation exists for the following sets of ordered pairs: {(-1, 1), (0, 3), (-1/3, 7/3)}. The rule for the linear equati

olya-2409 [2.1K]

The rule of the linear equation which is 'y is three more than twice x' can be translated into y = 2x + 3. Thus, we simply use the points and subtitute it to the y and x variables.

1st point: (-1, 1)
1 = 2(-1) + 3
1 = -2 + 3
1=1

2nd point: (0, 3)
3 = 2(0) + 3
3 =3

3rd point: (-1/3, 7/3)
7/3 = 2(-1/3) + 3
7/3 = -2/3 + 3
7/3 = 7/3

All points prove the equation satifsfies the set of coordinates.

8 0

4 years ago

Read 2 more answers

3(x+6)+6x+9=7x+8 solve for x

fomenos

Answer:

x=-19/2

Step-by-step explanation:

3(x+6)+6x+9=7x+8

3x+18+6x+9=7x+8

9x+27=7x+8

9x+27-7x=8

2x+27=8

2x=8-27

2x=-19

x=-19/2

7 0

3 years ago