How to find the slope (In general)

1 answer:

8 0

Using two of the points on the line, you can find the slope of the line by finding the rise and the run. The vertical change between two points is called the rise, and the horizontal change is called the run. The slope equals the rise divided by the run: Slope =riserun Slope = rise run .

Type of Slope: Visual Description

Undefined: vertical

Negative: downhill

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For questions 10-11 if l || m, find the values of x and y.

klemol [59]

Answer:

10) 9x - 2° = 5x + 54° (corresponding angles are equal)

9x - 5x = 54 + 2

4x = 56

x = 56/4

x = 14°

10y + 6° = 9x - 2° (linear pair)

10y + 6° = 9(14)° - 2°

10y + 6° = 126° - 2°

10y + 6° = 124°

10y = 124° - 6°

10y = 118

y = 118/10

Sorry, i don't know how to do the 11th question

but hope this helps you!

8 0

2 years ago

Riley rides his bicycle 1.8 km to Jillian’s house. On the way back, he takes a route that is 740 m shorter than the first route.

postnew [5]

1.8 km >> 1,800 m
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The shorter route is 1.06 km.

7 0

2 years ago

Read 2 more answers

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$