Question

Mrs. McDonald is going to buy coffee and tea for her local food pantry. At Costless she can buy a canister of coffee, c
, for $7 and a box of tea, b
, for $4. This is modeled by 7c+4b
. Evaluate for c=10
and b=8
to find how much it will cost Mrs. McDonald to purchase 10 canisters of coffee and 8 boxes of tea.(1 point)
Responses

$592
$592

$96
$96

$29
$29

$102

Answer 1

Simplify 79h + 10 − 49h − 3
.(1 point)
Responses

39h + 7 = 13h + 7
Start Fraction 3 over 9 End Fraction h plus 7 equals Start Fraction 1 over 3 End Fraction h plus 7

318h + 7 = 16h + 7
Start Fraction 3 over 18 End Fraction h plus 7 equals Start Fraction 1 over 6 End Fraction h plus 7

318h2 + 7 = 16h2 + 7
Start Fraction 3 over 18 End Fraction h squared plus 7 equals Start Fraction 1 over 6 End Fraction h squared plus 7

39h2 + 7 = 13h2 + 7

Answer 2

Simplify 7/9h + 10 − 4/9h − 3
.(1 point)
Responses

3/9h + 7 = 1/3h + 7
Start Fraction 3 over 9 End Fraction h plus 7 equals Start Fraction 1 over 3 End Fraction h plus 7

318h + 7 = 16h + 7
Start Fraction 3 over 18 End Fraction h plus 7 equals Start Fraction 1 over 6 End Fraction h plus 7

318h2 + 7 = 16h2 + 7
Start Fraction 3 over 18 End Fraction h squared plus 7 equals Start Fraction 1 over 6 End Fraction h squared plus 7

39h2 + 7 = 13h2 + 7

Answer 3

Simplify 38x−910y+38x+710y
.(1 point)
Responses

68x+210y=34x+15y
Start Fraction 6 over 8 End Fraction x plus Start Fraction 2 over 10 End Fraction y equals Start Fraction 3 over 4 End Fraction x plus Start Fraction 1 over 5 End Fraction y

68x−210y=34x−15y
Start Fraction 6 over 8 End Fraction x minus Start Fraction 2 over 10 End Fraction y equals Start Fraction 3 over 4 End Fraction x minus Start Fraction 1 over 5 End Fraction y

616x+220y=38x+110y
Start Fraction 6 over 16 End Fraction x plus Start Fraction 2 over 20 End Fraction y equals Start Fraction 3 over 8 End Fraction x plus Start Fraction 1 over 10 End Fraction y

616x−220y=38x−110y

Answer 4

Simplify 3/8x−9/10y+3/8x+7/10y
.(1 point)
Responses

6/8x+2/10y=3/4x+1/5y
Start Fraction 6 over 8 End Fraction x plus Start Fraction 2 over 10 End Fraction y equals Start Fraction 3 over 4 End Fraction x plus Start Fraction 1 over 5 End Fraction y

6/8x−2/10y=3/4x−1/5y
Start Fraction 6 over 8 End Fraction x minus Start Fraction 2 over 10 End Fraction y equals Start Fraction 3 over 4 End Fraction x minus Start Fraction 1 over 5 End Fraction y

6/16x+2/20y=3/8x+1/10y
Start Fraction 6 over 16 End Fraction x plus Start Fraction 2 over 20 End Fraction y equals Start Fraction 3 over 8 End Fraction x plus Start Fraction 1 over 10 End Fraction y

6/16x−2/20y=3/8x−1/10y

Answer 5

To find the total cost for Mrs. McDonald to purchase 10 canisters of coffee and 8 boxes of tea, we can use the formula given, which is:

\[ 7c + 4b \]

Where:
- \( c \) is the number of canisters of coffee,
- \( b \) is the number of boxes of tea.

Substituting \( c = 10 \) and \( b = 8 \) into the equation:

\[
7(10) + 4(8)
\]

Calculating each term:

1. \( 7(10) = 70 \)
2. \( 4(8) = 32 \)

Now, add these two results together:

\[
70 + 32 = 102
\]

Therefore, the total cost for Mrs. McDonald to purchase 10 canisters of coffee and 8 boxes of tea is **$102**.

The correct response is:

**$102**

Answer 6

To simplify the expression \( 79h + 10 - 49h - 3 \), we can follow these steps:

1. Combine the like terms involving \( h \):
\[
79h - 49h = 30h
\]

2. Combine the constant terms:
\[
10 - 3 = 7
\]

Putting it all together, we get:
\[
30h + 7
\]

Now let's compare this result with the options provided:

1. \( 39h + 7 = 13h + 7 \)
2. \( \frac{3}{9}h + 7 = \frac{1}{3}h + 7 \)
3. \( \frac{3}{18}h^2 + 7 = \frac{1}{6}h^2 + 7 \)
4. \( 39h^2 + 7 = 13h^2 + 7 \)

None of these responses match the simplified expression \( 30h + 7 \). It appears there might have been an error in the options provided or in my calculations.

If the task was simply to simplify \( 79h + 10 - 49h - 3 \), the correct simplified expression is:

**\( 30h + 7 \)**.

Answer 7

To simplify the expression \( \frac{7}{9}h + 10 - \frac{4}{9}h - 3 \), we will follow these steps:

1. **Combine like terms** (the terms with \( h \)):
\[
\frac{7}{9}h - \frac{4}{9}h = \frac{7 - 4}{9}h = \frac{3}{9}h
\]

2. **Combine the constant terms**:
\[
10 - 3 = 7
\]

Putting it all together, the expression simplifies to:
\[
\frac{3}{9}h + 7
\]

3. **Further simplify \(\frac{3}{9}h\)**:
\[
\frac{3}{9}h = \frac{1}{3}h
\]

So we can rewrite the final expression as:
\[
\frac{1}{3}h + 7
\]

Now, comparing with the options provided:

1. \( \frac{3}{9}h + 7 = \frac{1}{3}h + 7 \)
2. \( \frac{3}{18}h + 7 = \frac{1}{6}h + 7 \)
3. \( \frac{3}{18}h^2 + 7 = \frac{1}{6}h^2 + 7 \)
4. \( 39h^2 + 7 = 13h^2 + 7 \)

The correct option that matches our simplified expression is:

**\( \frac{3}{9}h + 7 = \frac{1}{3}h + 7 \)**.

Answer 8

Let's simplify the expression \( 38x - 910y + 38x + 710y \).

1. **Combine the like terms** involving \( x \):
\[
38x + 38x = 76x
\]

2. **Combine the like terms** involving \( y \):
\[
-910y + 710y = -200y
\]

Putting it all together, we have:
\[
76x - 200y
\]

Now, let's look through the provided options to see if any match our simplified expression:

1. \( 68x + 210y = 34x + 15y \)
2. \( 68x - 210y = 34x - 15y \)
3. \( 616x + 220y = 38x + 110y \)
4. \( 616x - 220y = 38x - 110y \)

None of these options match our simplified expression \( 76x - 200y \).

### Conclusion
The final simplified expression is:
\[
76x - 200y
\]

Please let me know if there are specific conditions or modifications you would like reflected in the simplification process!

Answer 9

To simplify the expression \( \frac{3}{8}x - \frac{9}{10}y + \frac{3}{8}x + \frac{7}{10}y \), we will follow these steps:

1. **Combine the like terms** involving \( x \):
\[
\frac{3}{8}x + \frac{3}{8}x = \frac{6}{8}x
\]

2. **Combine the like terms** involving \( y \):
\[
-\frac{9}{10}y + \frac{7}{10}y = -\frac{2}{10}y
\]

Putting it all together, we get:
\[
\frac{6}{8}x - \frac{2}{10}y
\]

Next, we can simplify the coefficients further:

- \( \frac{6}{8}x \) simplifies to \( \frac{3}{4}x \) (dividing the numerator and the denominator by 2).
- \( -\frac{2}{10}y \) simplifies to \( -\frac{1}{5}y \) (dividing the numerator and the denominator by 2).

Thus, the simplified expression can also be written as:
\[
\frac{3}{4}x - \frac{1}{5}y
\]

Now, let's compare this with the provided options:

1. \(\frac{6}{8}x + \frac{2}{10}y = \frac{3}{4}x + \frac{1}{5}y\)
2. \(\frac{6}{8}x - \frac{2}{10}y = \frac{3}{4}x - \frac{1}{5}y\)
3. \(\frac{6}{16}x + \frac{2}{20}y = \frac{3}{8}x + \frac{1}{10}y\)
4. \(\frac{6}{16}x - \frac{2}{20}y = \frac{3}{8}x - \frac{1}{10}y\)

From the simplified expression \( \frac{6}{8}x - \frac{2}{10}y \), we can directly see that this matches option 2:

**\(\frac{6}{8}x - \frac{2}{10}y = \frac{3}{4}x - \frac{1}{5}y\)**.