Nine negative third power in standard form

Consider the expression 17x+8x+35.Which of these mathematical statements could be used to accurately describe the given expressi

Mrac [35]

3. The product of seventeen and a number, increased by the product of eight and the same number, increased by thirty/five.
4. Seventeen times a number plus eight times the same number plus thirty five

5 0

3 years ago

AB, CD, and EF intersect at O. If CD is perpendicular to EF and m

REY [17]

Answer:

m(∠AOF) = 148°

Step-by-step explanation:

From the figure attached,

CD intersects line EF at a point O.

Line CD is perpendicular to the line EF.

m(∠AOE) = 32°

m(∠COE) = 90°

Since m(∠COE) = m(∠AOE) + m(∠AOC) = 90°

32° + m(∠AOC) = 90°

m(∠AOC) = 90° - 32° = 58°

m(∠AOF) = m(∠AOC) + m(∠COF)

= 58° + 90°

= 148°

Therefore, m(∠AOF) = 148° will be the answer.

4 0

3 years ago

What expressions are equivalent to 7^3 * 7^x

solong [7]

Answer:

The question = (7^3)*(7^x)

When multiplying two exponents with the same base, all you do is add the exponents together and keep the same base

Therefore,

(7^3)*(7^x) = 7^(3+x)

Step-by-step explanation:

7^3+x is the answer

3 0

3 years ago

Help please!!!!!!!!!

In-s [12.5K]

ANSWER

24

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}$

so $C_{11} = M_{11} = 24$

4 0

3 years ago

7×9-8+11-6 [5+2 (-3)]

kupik [55]

Ok, MissWalker! Please try your best to understand this, it may get confusing.
I have solved this on a different website, and here is my solving in a picture. I hope I helped!!! :D

8 0

3 years ago