Select all of the following that are like radicals. 9./6ab2 -5./6ab2 436ab2 2 V6b²a

Simplify the expression 5-2(9)|+9^2÷3

Delicious77 [7]

Answer:

14

Step-by-step explanation:

Simplify the following:

5 - 2×9 + 9^2/3

Hint: | Evaluate 9^2.

9^2 = 81:

5 - 2×9 + 81/3

Hint: | Reduce 81/3 to lowest terms. Start by finding the GCD of 81 and 3.

The gcd of 81 and 3 is 3, so 81/3 = (3×27)/(3×1) = 3/3×27 = 27:

5 - 2×9 + 27

Hint: | Multiply -2 and 9 together.

-2×9 = -18:

5 + -18 + 27

Hint: | Evaluate 5 + 27 using long addition.

| 1 |

| 2 | 7

+ | | 5

| 3 | 2:

32 - 18

Hint: | Subtract 18 from 32.

| 2 | 12

| 3 | 2

- | 1 | 8

| 1 | 4:

Answer: 14

5 0

3 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

Trigonometric Identities

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

$\sin(\angle ABC)=\dfrac{4}{5}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$

7 0

2 years ago

Read 2 more answers

I put B but it said it was wrong???

zavuch27 [327]

It should be A. Not B

8 0

3 years ago

Guadalupe is making a project for shop class. He has four pieces of

ipn [44]

Answer: The piece of wood will be 23.2 in. long

Step-by-step explanation:

5.8 multiplied by 4 is 23.2

8 0

3 years ago

What is 4C (matrix) ?

Cerrena [4.2K]

The 4C matrix is obtained when all elements of matrix C are multiplied by 4. It has all the same elements as C, only multiplied by 4.
Let's analyze all matrices
A. Not all elements of A can be divided by 4 (3 can not be divided by 4, the solution is not a whole number).
B. All elements of matrix B can be divided by 4, which means that B is a $C matrix (there is a matrix C which multiplied by 4 gives the matrix B).
C. Not all elements of C can be divided by 4.
D. Not all elements of D can be divided by 4.

Solution: B

4 0

3 years ago