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d1i1m1o1n [39]
2 years ago
15

An adventure crew takes a hike in the woods. The path of their hike is shown below. Write an expression to represent the total d

istance of the hike and then factor the expression. Side a is 4x+1 side b:5x+4 side c:6x+5 I need the total distance and the factor of the equation this is factoring by gcf
Mathematics
1 answer:
Sliva [168]2 years ago
7 0

The total distance covered during hiking is 15x + 10

<h3>What is distance?</h3>

Distance is the amount of space between two points.

Given that  Side a is 4x+1 side b:5x+4 side c:6x+5, hence:

Total distance = side a + side b + side c

Total distance = 4x + 1 + 5x + 4 + 6x + 5 = 15x + 10

The factor of the total distance is:

15x + 10 = 0

15x = -10

x = -2/3

The total distance covered during hiking is 15x + 10

Find out more on distance at: brainly.com/question/17273444

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Please help me to prove this!​
Sophie [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C                → A = C - B

                                          → B = C - A

Use the Double Angle Identity:     cos 2A = 2 cos² A - 1

                                             → (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS:                     cos² A + cos² B + cos² C

\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C

\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C

\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C

\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]

\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)

\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)

\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C   \checkmark

5 0
3 years ago
If f(x) = x + 15, find f(3)
gavmur [86]

Answer:

I think its 18

Step-by-step explanation:

3 0
3 years ago
What is the slope of a line that is parallel to y=5x + 3
Sedaia [141]
Hi there!

The slope of this line is 5*. Therefore, the slope of any line that is parallel to y = 5x + 3 is 5 as well (since parallel lines have the same slope).

* In a linear formula in standard form y = ax + b the a represents the slope of the line. In other words: in a linear formula in above form, the slope is the number in front of x.
6 0
4 years ago
Read 2 more answers
Someone plz help me :(
ivanzaharov [21]

Answer:

B) 5 gallons

Step-by-step explanation:

4 quarts = 1 gallon

multiply each number by 5

20 quarters = 5 gallons

4 0
3 years ago
Read 2 more answers
The question is above
galina1969 [7]
For 11, we can subtract x from both sides for the first equation to get y=8-x, making the slope -1 for both equations and therefore making it parallel.

For 13, we can solve for y in the first equation by subtracting 14 from both sides and therefore getting 2x-14=y. For the next one, we can subtract 4x from both sides to get 10-4x=2y. Dividing both sides by 2, we get 10-2x=y. Since for the equations to be parallel they must be the same and for perpendicular they must be -1/(slope), this fits neither.

Using those two explanations, I implore you to solve the other two on your own!
3 0
3 years ago
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