Use the identities for the cosine of a sum or a difference to write the expression as a single function of x.

I’m stuck I need the answer

Mama L [17]

This is the answer for your test hope it helps

8 0

2 years ago

Simplify the expression below. 12b + 86 + 5b 0 25 + b O 25(3b) 25b 3b

Alja [10]

(1): "^-5" was replaced by "^(-5)". 3 more similar replacement(s)

STEP

1

:

Equation at the end of step 1

((25•(b-1))•(c-3))

(((12•(b4))•(c-6))•——————————————————)•((2•5b(-5))•c)

((15•(b3))•(c4))

STEP

2

:

Equation at the end of step

2

:

((25•(b-1))•(c-3))

(((12•(b4))•(c-6))•——————————————————)•(2•5b(-5)c)

((3•5b3)•c4)

STEP

3

:

Equation at the end of step

3

:

(52b(-1)•c(-3))

(((12•(b4))•(c-6))•———————————————)•(2•5b(-5)c)

(3•5b3c4)

STEP

4

:

52b(-1)c(-3)

Simplify ————————————

(3•5b3c4)

Dividing exponential expressions :

4.1 b(-1) divided by b3 = b((-1) - 3) = b(-4) = 1/b4

Dividing exponential expressions :

4.2 c(-3) divided by c4 = c((-3) - 4) = c(-7) = 1/c7

Dividing exponents:

4.3 52 divided by 51 = 5(2 - 1) = 51 = 5

Equation at the end of step

4

:

5

(((12•(b4))•(c-6))•—————)•(2•5b(-5)c)

3b4c7

STEP

5

:

Equation at the end of step

5

:

5

(((22•3b4) • c(-6)) • —————) • (2•5b(-5)c)

3b4c7

STEP

6

:

Canceling Out :

6.1 Canceling out b4 as it appears on both sides of the fraction line

Dividing exponential expressions :

6.2 c(-6) divided by c7 = c((-6) - 7) = c(-13) = 1/c13

Canceling Out:

6.3 Canceling out 3 as it appears on both sides of the fraction line

Equation at the end of step

6

:

20

——— • (2•5b(-5)c)

c13

STEP

7

:

Multiplying exponents:

7.1 22 multiplied by 21 = 2(2 + 1) = 23

Multiplying exponents:

7.2 51 multiplied by 51 = 5(1 + 1) = 52

Dividing exponential expressions :

7.3 c1 divided by c13 = c(1 - 13) = c(-12) = 1/c12

Final result :

200

—————

b5c12 Answer:

Step-by-step explanation:

5 0

2 years ago

what is the quotient (65y3 15y2 − 25y) ÷ 5y? a. 13y2 3y − 5 b. 13y3 3y2 − 5y c. 13y2 − 3y 5 d.13y2 − 3y − 5

Aneli [31]

Keywords:

Division, quotient, polynomial, monomial

For this case we must solve a division between a polynomial and a monomial and indicate which is the quotient.

By definition, if we have a division of the form: $\frac {a} {b} = c$ , the quotient is given by "c".

We have the following polynomial:

$65y ^ 3 + 15y ^ 2 - 25y$ that must be divided between monomy $5y$ , then:

$C (y)$ represents the quotient of the division:

$C (y) = \frac {65y ^ 3 + 15y ^ 2 - 25y} {5y}$

$C (y) = \frac {65y ^ 3} {5y} + \frac {15y ^ 2} {5y} - \frac {25y} {5y}$

$C (y) = 13y ^ 2 + 3y-5$

Thus, the quotient of the division between the polynomial and the monomial is given by:

$C (y) = 13y ^ 2 + 3y-5$

Answer:

The quotient is: $C (y) = 13y ^ 2 + 3y-5$

Option: A

4 0

2 years ago

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BRAINLIEST Please help ASAP and leave a small explanation :)

Andreas93 [3]

1 and 8 are because they are alternate exterior angles

3 0

3 years ago

Triangle T R S has centroid Z. Lines are drawn from each point to the midpoint of the opposite side to form line segments T W, R

solong [7]

Answer:

C?

Step-by-step explanation:

From eyeballing it, I can see that RZ is definetely not the first two. and RZ looks about exactly double the length of VZ. Hope this helps. :)

6 0

2 years ago

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