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3 years ago

Write an algebraic expression, 1 less than the quotient of a number n and 6

Mathematics

2 answers:

spayn [35]3 years ago

7 0

" the quotient " means divide

(n/6) - 1 <== ur expression

sashaice [31]3 years ago

4 0

I think its 1 < n/6
im new so if it right i would be happy xD

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Which is the graph of y=2/(x+1)-6

krek1111 [17]

Answer:

I have linked the graph in an image

Step-by-step explanation:

1. Simplify y=2/(x+1)-6

2. Subtract the numbers 1-6=-5

3. = 2/x-5

4. Graph your results

4 0

2 years ago

Read 2 more answers

Question 1 of 12, Step 1 of 1

stellarik [79]

Answer:

325 and 375 mph

Step-by-step explanation:

Given :

Distance between plane A and B = 2800

Recall :

Distance = speed * time

Let speed of A = v1

Speed of B = v2

v1 = v2 + 50

Time taken = 4 hours

Distance = speed * time

Total distance = 2800

2800 = v1 * 4 + v2 * 4

2800 = (4v1) + (4v2)

2800 = 4(v2+50) + 4v2

2800 = 4v2 + 200 + 4v2

2800 = 8v2 + 200

2800 - 200 = 8v2

2600 = 8v2

v2 = 2600/8

v2 = 325 mph

v1 = v2 + 50

v1 = 325 + 50

v1 = 375 mph

8 0

2 years ago

A stack of bricks is 3 1/2 feet high.

ololo11 [35]

Divide 42 inches by the total amount of inches in one brick
why you are dividing by 42 inches is because 3 1/2 ft is equal to 42 inches

4 0

3 years ago

HELP ME ASAP on these math question it’s due after class

lana66690 [7]

I thin the answer is B

8 0

2 years ago

Please prove this........

Crazy boy [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago