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ddd [48]
2 years ago
11

How to find the slope (In general)

Mathematics
1 answer:
alexira [117]2 years ago
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
1,800 - 740 = 1,060 m
1,060 m >> 1.06 km
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.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.

From equation (i), we get

\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.

Substituting the values of A and B in the general solution, we get

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

Thus, the required solution of the given IVP is

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

4 0
3 years ago
What is 3 2/5 minus 1 4/5?
umka21 [38]
You can convert that to 17/5 - 9/5 which equals 8/5
4 0
2 years ago
PLEASE ANSWERR!
Zielflug [23.3K]

Answer:

1. is 5x+12 hope it helped !

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
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