All categories

2 years ago

What is the answer to this problem?

Mathematics

1 answer:

Alexeev081 [22]2 years ago

6 0

Answer:

gggggykkkk i just stole your points thank you

You might be interested in

Yasmin purchased 6 heads of cabbage that each weighed 2 and StartFraction 3 over 8 EndFraction pounds. How much did the cabbage

padilas [110]

The total weight of cabbage is 14 1/4

<h3 /><h3>What is Equation?</h3>

An equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value.

Given:

Yasmin purchased 6 heads of cabbage that each weighed 2 and 3/8 pounds.

So,

2 and 3/8 = 19/8

1 cabbage = 19/8 pounds

6 cabbages = x pounds

1 * x = 6 * 19/8

x = 114/8

x = 14 2/8

x = 14 1/4

Learn more about this concept here:

brainly.com/question/18762466

#SPJ1

4 0

1 year ago

Find the equation of the line.<br> Use exact numbers.<br> y =<br> X +

puteri [66]

Answer:

$y = \frac{3}{4} x - 2$

Step-by-step explanation:

Remmeber this:

$slope = \frac{rise}{run}$

The y-intercept is where the lone goes through the y-axis.

5 0

2 years ago

Captain Ishaan has a ship, the H.M.S. Khan. The ship is two furlongs from the dread pirate Luis and his merciless

fredd [130]

Answer:

Khan Academy's answer is 4/35.

Step-by-step explanation:

4/5 * 1/7 = 4/35

4 0

2 years ago

What’s the answer???

Over [174]

Answer:

Question 7 = $2

Question 8 = $16

8 0

2 years ago

Read 2 more answers

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

2 years ago