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3 years ago

Thaw relationship between a distance in yards and the same distance in miles is described by the equation

Mathematics

1 answer:

balandron [24]3 years ago

3 0

Answer:

The relationship between a distance in yards (y) and the same distance in miles (m) is described by the equation y = 1760m.

Step-by-step explanation:

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Sedaia [141]

Answer:

Margaret has 2 flowers

Step-by-step explanation:

First we can make a set of equations

j = 3m

a = 3j

a + j + m = 26

Since j = 3m we can rewrite Anna's equation as a = (3(3m)) or simply a = 9m

Now we can plug in the equations for Anna and Jessica into the main equation

9m + 3m + m = 26

13m = 26

m = 2

6 0

2 years ago

Which inequality is represented by this graph?

Arada [10]

Answer:

It C just trust me I swear

6 0

3 years ago

Mind helping me if so thank you

icang [17]

hed have to write 24 pages per hour to get up to 48 pages this weak.

5 0

3 years ago

Read 2 more answers

The left and right page numbers of an open book are two consecutive integers whose sum is 321

PIT_PIT [208]

Answer:

160,160

Step-by-step explanation:

let the first page be X

second one be x+1

x+x+1=321

2x=320

x= 320÷2=160

pages 160 161

6 0

3 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