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goblinko [34]
3 years ago
13

real brought 3 college textbooks one cost $32 one cost $45 one cost $39 what is the average price of his books

Mathematics
2 answers:
taurus [48]3 years ago
8 0
$40 is the average price i might be wrong but i am pretty sure
vodomira [7]3 years ago
3 0
(32 + 45 + 39) \div 3

The answer would be $38.67
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Help, ASAP! Please!!
Yuliya22 [10]

Answer:

a. Parallel

b. Not parallel

Step-by-step explanation:

Systems which have no solutions are parallel lines. These lines do not intersect and therefore have no solution. Remember, parallel lines have the same slope. Compare the slope in each equation when in y=mx+b format to see if they are parallel.

a. y= 2x + 3                                                  y-2x=-3 becomes y = 2x - 3

m = 2                                                            m= 2

The slopes are the same so these are parallel lines.

b. 3x + y = 2 becomes y= -3x + 2               y = 1/3 x + 1/2

m = -3                                                          m= 1/3

The slopes are different so this is not parallel.

Since -3 and 1/3 are the negative reciprocals of each other. These lines are actually perpendicular.

7 0
3 years ago
Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A<br>What is the value of the A^2?<br><br>​
Alla [95]

\large \mathbb{PROBLEM:}

\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}

\large \mathbb{SOLUTION:}

\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}

\boxed{ \tt   \red{C}arry  \: \red{ O}n \:  \red{L}earning}  \:  \underline{\tt{5/13/22}}

4 0
2 years ago
There are 3 consecutive even integers that sum to 24. What is the value of the greatest integer?
Hunter-Best [27]

Answer:

10

Step-by-step explanation:

x+x+2+x+4=24

3x+6 = 24

x=6

x+4=10

7 0
3 years ago
Read 2 more answers
Suppose that the p-value was 0.0259. what is the appropriate conclusion to make if α = 0.05?
fiasKO [112]
<span>It is clear to see that 0.0259 is less than 0.05. A p-value that is less than the confidence level of alpha indicates that there is sufficient evidence from the data that the null hypothesis should be rejected in favor of an alternative hypothesis.</span>
6 0
3 years ago
Based on Pythagorean identities, which equation is true? A. Sin^2 theta -1= cos^2 theta B. Sec^2 theta-tan^2 theta= -1 C. -cos^2
Arturiano [62]

Answer:

D

Step-by-step explanation:

our basic Pythagorean identity is cos²(x) + sin²(x) = 1

we can derive the 2 other using the listed above.

1. (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

1 + tan²(x) = sec²(x)

2.(cos²(x) + sin²(x))/sin²(x) = 1/sin²(x)

cot²(x) + 1 = csc²(x)

A. sin^2 theta -1= cos^2 theta

this is false

cos²(x) + sin²(x) = 1

isolating cos²(x)

cos²(x) = 1-sin²(x), not equal to sin²(x)-1

B. Sec^2 theta-tan^2 theta= -1

1 + tan²(x) = sec²(x)

sec²(x)-tan(x) = 1, not -1

false

C. -cos^2 theta-1= sin^2

cos²(x) + sin²(x) = 1

sin²(x) = 1-cos²(x), our 1 is positive not negative, so false

D. Cot^2 theta - csc^2 theta=-1

cot²(x) + 1 = csc²(x)

isolating 1

1 = csc²(x) - cot²(x)

multiplying both sides by -1

-1 = cot²(x) - csc²(x)

TRUE

3 0
3 years ago
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