All categories

3 years ago

Which equation is an identity.

Mathematics

1 answer:

Verizon [17]3 years ago

6 0

for 1 the equation is an identity is A. number 2 the equation has no solution is D.

You might be interested in

Question 12 please , thank you in advance.

Monica [59]

I thing it might be C

5 0

4 years ago

Read 2 more answers

On line l, m∠p = 21x +25 and m∠n = \{675x\:-27}/3 Use this information to answer Part A & B below. Part A: Determine the val

tensa zangetsu [6.8K]

Answer:

a) x = 492/738

b) m<p = 39°

m<n = 141°

Step-by-step explanation:

Find the diagram attached

a) The sum of angle on the straight line l is 180degrees

m<n + m<p = 180 degrees

Given

<p = 21x+25

<n = (675x-27)/3

Substitute the given values into the formula

(675x-27)/3 + 21x+25= 180

[(675x-27+3(21x+25))]/3 = 180

675x-27+63x +75 = 180*3

675x+63x-27+75 = 540

738x+48 = 540

738x = 540 - 48

738x = 492

x = 492/738

b) m<p = 21x+25

m<p = 21(492/738) + 25

m<p = 14+25

m<p = 39°

Recall that m<p + m<n = 180

m<n = 180-m<p

m<n = 180 -39

m<n = 141°

4 0

3 years ago

The original quantity is 10 and the new quantity is 13. What is the percent change?

Leya [2.2K]

3 is 13 more than 10, and 3 is 30% of 10, so 30% im pretty sure

6 0

3 years ago

Read 2 more answers

The revenue equation (in hundreds of millions of dollars) for barley production in a certain country is approximated by R(x) = 0

STatiana [176]

Answer:

The marginal-revenue equation is $R'(x)=0.1226x+1.1584$ .

The marginal revenue for the production of 200,000,000 bushels is $R'(x)=1.4036$ .

The marginal revenue for the production of 650,000,000 bushels is $R'(x)=9.1274$ .

Step-by-step explanation:

The derivative $R'(x)$ of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.

We know the revenue function, $R(x) = 0.0613x^{2} +1.1584x+2.4813$ . Therefore, the marginal revenue function is

$R'(x)=\frac{d}{dx}R(x) = \frac{d}{dx}(0.0613x^{2} +1.1584x+2.4813)\\\\=\frac{d}{dx}\left(0.0613x^2\right)+\frac{d}{dx}\left(1.1584x\right)+\frac{d}{dx}\left(2.4813\right)\\\\R'(x)=0.1226x+1.1584$