Which property is illustrated by the equation below: 6(5+1)=6(5)+6(1)

Evaluate the expression 2ab 3+ 6a 3 - 4ab2 when a=2 and b=4. a.176 b.180 c.215 d.216

Tom [10]

The first step for solving this expression is to insert what a and b stand for into the expression. This will change the expression to the following:
2(2)(4)³ + 6(2)³ - 4(2)(4)²
Now we can start solving this by factoring the expression
2(2 × 4³ + 3 × 2³ - 4 × 4²)
Write 4³ in exponential form with a base of 2.
2(2 × $2^{6}$ + 3 × 2³ - 4 × 4²)
Calculate the product of -4 × 4².
2(2 × $2^{6}$ + 3 × 2³ -4³)
Now write 4³ in exponential form with a base of 2.
2(2 × $2^{6}$ + 3 × 2³ - $2^{6}$ )
Collect the like terms with a base of 2.
2( $2^{6}$ + 3 × 2³)
Evaluate the power of 2³.
2( $2^{6}$ + 3 × 8)
Evaluate the power of $2^{6}$ .
2(64 + 3 × 8)
Multiply the numbers.
2(64 + 24)
Add the numbers in the parenthesis.
2 × 88
Multiply the numbers together to find your final answer.
176
This means that the correct answer to your question is option A.
Let me know if you have any further questions.
:)

5 0

3 years ago

Two of the lights at the local stadium are flickering. They both just flickered at the same time.

V125BC [204]

Answer:

76 seconds

Step-by-step explanation:

You need to find the first common multiple of both 7 and 8. the LCM is actually the numbers multiplies by each other, or 56.

4 0

3 years ago

The monthly revenue, f(x) for the Sweet Treats Ice Cream truck

irakobra [83]

Answer:

The vertex is (-4, -52), and it represents the minimum for the function.

Step-by-step explanation:

3 0

3 years ago

What eigen value for this matix (1 -2) (-2 0)

natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

Compute the matrix $A' = A-\lambda I$ , where I is the identity matrix (1s on the diagonal, 0s elsewhere)
Compute the determinant of A'
Set the determinant of A' equal to zero and solve for lambda.

So, in this case, we have

$A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]$