What is the rate of change and initial value for the linear relation that includes the points shown in the table?

At a restaurant, three salads, two sandwiches, and one drink cost $17.75. One salad, one sandwich, and three drinks cost $10.75.

Leno4ka [110]

Answer:

A salad costs $2.50

A sandwich costs $4.50

A drink costs $1.25

Step-by-step explanation:

Let x represent the salad, y represent the sandwich, and z represent the drink.

Since three salads, two sandwiches, and one drink cost $17.75:

3x + 2y + z = 17.75 (1)

Since one salad, one sandwich, and three drinks cost $10.75:

x + y + 3z = 10.75 (2)

Since a salad costs twice as much as a drink:

x = 2z (3)

Multiply the equation 2 by -2 then, sum the equation 1 and equation 2:

-2x - 2y - 6z = -21.50

3x + 2y + z = 17.75

→ x - 5z = -3.75

Replace the x with 2z using equation 3:

2z - 5z = -3.75

-3z = -3.75

z = 1.25

x = 2z → x = 2.50

x + y + 3z = 10.75 → 2.50 + y + 3.75 = 10.75 → y + 6.25 = 10.75 → y = 4.50

5 0

2 years ago

2. Sequoia trees can grow to more than 90 m in

sashaice [31]

Answer: F

Step-by-step explanation: 90/3.28 equals 27.4390244

5 0

2 years ago

) f) 1 + cot²a = cosec²a

notsponge [240]

Answer:

It is an identity, proved below.

Step-by-step explanation:

I assume you want to prove the identity. There are several ways to prove the identity but here I will prove using one of method.

First, we have to know what cot and cosec are. They both are the reciprocal of sin (cosec) and tan (cot).

$\displaystyle \large{\cot x=\frac{1}{\tan x}}\\\displaystyle \large{\csc x=\frac{1}{\sin x}}$

csc is mostly written which is cosec, first we have to write in 1/tan and 1/sin form.

$\displaystyle \large{1+(\frac{1}{\tan x})^2=(\frac{1}{\sin x})^2}\\\displaystyle \large{1+\frac{1}{\tan^2x}=\frac{1}{\sin^2x}}$

Another identity is:

$\displaystyle \large{\tan x=\frac{\sin x}{\cos x}}$

Therefore:

$\displaystyle \large{1+\frac{1}{(\frac{\sin x}{\cos x})^2}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{1}{\frac{\sin^2x}{\cos^2x}}=\frac{1}{\sin^2x}}\\\displaystyle \large{1+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}$

Now this is easier to prove because of same denominator, next step is to multiply 1 by sin^2x with denominator and numerator.

$\displaystyle \large{\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}}\\\displaystyle \large{\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$