Question

assignment
name
date 12
period
evaluate each expression.

1. (1 + 5) × 6
2. (3 + 5) + 2

simplify each expression.

1. -7x + x
2. r - 1 - 3
3. -9p - 2p
4. 8n - 9 + 3 + 6n

simplify. your answer should contain only positive exponents.

1. m³ · 5m³
2. 6x⁴ · 8x
3. for washing the dog scott was given $16.96. now he has $45.76. how much money did he have before?
4. imani won 55 super bouncy balls playing horseshoes at the county fair. at school she gave one to every student in her math class. she only has 43 remaining. how many did she give away?

evaluate each expression.

1. 2 + 1 - (5 - 4)
2. (4 - 1) × 3 × 5
3. scott and his best friend found some money buried in a field. they split the money evenly, each getting $8. how much money did they find?
4. shanice spent $36 on cereal. if they cost $6 / box, how many boxes did she buy?

solve each equation.

1. -8 = x - 19
2. -14 + b = -21

Answer:

Let's solve these problems one by one. We'll start with the first few:

Problem 1: Evaluate \((1 + 5) \times 6\)

Step 1: Simplify inside the parentheses

First, we calculate the sum inside the parentheses: \(1 + 5 = 6\).

Step 2: Multiply the result by 6

Now we multiply that result by 6: \(6 \times 6 = 36\).
So the answer for problem 1 is \(36\).

Problem 2: Evaluate \((3 + 5) \div 2\)

Step 1: Simplify inside the parentheses

Calculate the sum inside the parentheses: \(3 + 5 = 8\).

Step 2: Divide the result by 2

Now divide that result by 2: \(8 \div 2 = 4\).
So the answer for problem 2 is \(4\).

Problem 3: Simplify \(-7x + x\)

Step 1: Combine like terms

We have \(-7x\) and \(x\) (which is \(1x\)). To combine them, we add their coefficients: \(-7 + 1 = -6\).
So \(-7x + x = -6x\).

Problem 4: Simplify \(r - 1 - 3\)

Step 1: Combine the constant terms

We have \(-1\) and \(-3\). Adding those together: \(-1 - 3 = -4\).
So \(r - 1 - 3 = r - 4\).

Problem 5: Simplify \(-9p - 2p\)

Step 1: Combine like terms

Add the coefficients of \(p\): \(-9 - 2 = -11\).
So \(-9p - 2p = -11p\).

Problem 6: Simplify \(8n - 9 + 3 + 6n\)

Step 1: Combine like terms for \(n\)

We have \(8n\) and \(6n\). Adding their coefficients: \(8 + 6 = 14\), so \(8n + 6n = 14n\).

Step 2: Combine the constant terms

We have \(-9\) and \(3\). Adding those: \(-9 + 3 = -6\).
So putting it together: \(14n - 6\).

Problem 7: Simplify \(m^3 \cdot 5m^3\) (with positive exponents)

Step 1: Multiply the coefficients

The coefficients are \(1\) (for \(m^3\)) and \(5\). Multiplying them: \(1 \times 5 = 5\).

Step 2: Multiply the variables using exponent rules

When multiplying variables with exponents, we add the exponents: \(m^3 \cdot m^3 = m^{3 + 3} = m^6\).
So putting it together: \(5m^6\).

Problem 8: Simplify \(6x^4 \cdot 8x\) (with positive exponents)

Step 1: Multiply the coefficients

\(6 \times 8 = 48\).

Step 2: Multiply the variables using exponent rules

\(x^4 \cdot x = x^{4 + 1} = x^5\) (since \(x = x^1\)).
So putting it together: \(48x^5\).

Problem 9: Scott's money before washing the dog

Let \(x\) be the money Scott had before. After getting \$16.96, he has \$45.76. So the equation is:
\(x + 16.96 = 45.76\)

Step 1: Solve for \(x\)

Subtract \(16.96\) from both sides: \(x = 45.76 - 16.96\).

Step 2: Calculate the difference

\(45.76 - 16.96 = 28.8\).
So Scott had \$28.80 before.

Problem 10: Bouncy balls Imani gave away

Let \(g\) be the number of balls given away. She started with 55 and has 43 left. So:
\(55 - g = 43\)

Step 1: Solve for \(g\)

Subtract 43 from both sides: \(55 - 43 = g\).

Step 2: Calculate the difference

\(55 - 43 = 12\).
So Imani gave away 12 balls.

Problem 11: Evaluate \(2 + 1 - (5 - 4)\)

Step 1: Simplify inside the parentheses

\(5 - 4 = 1\).

Step 2: Perform the addition and subtraction from left to right

\(2 + 1 = 3\), then \(3 - 1 = 2\).
So the answer is \(2\).

Problem 12: Evaluate \((4 - 1) \times 3 \times 5\)

Step 1: Simplify inside the parentheses

\(4 - 1 = 3\).

Step 2: Multiply the results

\(3 \times 3 = 9\), then \(9 \times 5 = 45\).
So the answer is \(45\).

Problem 13: Money Scott and his friend found

Let \(m\) be the total money found. They split it evenly, each getting \$8, and there are 2 people (Scott and his friend). So:
\(\frac{m}{2} = 8\)

Step 1: Solve for \(m\)

Multiply both sides by 2: \(m = 8 \times 2\).

Step 2: Calculate the product

\(8 \times 2 = 16\).
So they found \$16.

Problem 14: Boxes of cereal Shanice bought

Let \(b\) be the number of boxes. Each box costs \$6, and she spent \$36. So:
\(6b = 36\)

Step 1: Solve for \(b\)

Divide both sides by 6: \(b = \frac{36}{6}\).

Step 2: Calculate the quotient

\(\frac{36}{6} = 6\).
So Shanice bought 6 boxes.

Problem 15: Solve \(-8 = x - 19\)

Step 1: Isolate \(x\)

Add 19 to both sides: \(-8 + 19 = x\).

Step 2: Calculate the sum

\(-8 + 19 = 11\).
So \(x = 11\).

Problem 16: Solve \(-14 + b = -21\)

Step 1: Isolate \(b\)

Add 14 to both sides: \(b = -21 + 14\).

Step 2: Calculate the sum

\(-21 + 14 = -7\).
So \(b = -7\).