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vladimir1956 [14]
3 years ago
7

Pls helpppp me I’m super stuckkk

Mathematics
1 answer:
weeeeeb [17]3 years ago
5 0

Answer:

slope=2

Step-by-step explanation:

subrtract -4-(-4) the two negatives make it a + so that becomes 8 and thats for the top one the bottom is 6-2 which is 4 then divide 8 by 4 to find the slope which is 2.

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How do you do this simultaneous equation?
tatiyna
The y is the same for both equations. Equate them
-x - 5 = 2x + 4 Add x to both sides
- 5 = 3x + 4 Subtract 4 from both sides.
- 9 = 3x Divide by 3
x = -9 / 3
x = - 3

y = - x - 5
y = -(-3) - 5
y = - 2

So that point is where the two equations meet.

To see the graph of your question on the internet, go to 
Wolframalpha.com 
and put this string in the question space.
y = -x - 5; y  = 2x + 4  Note the semi colon in the middle.

To the moderator: This is my own work. I'm using Wolfram to get the answer.
5 0
3 years ago
Explain how you might go from gallons to cups using more than one conversion factor.
damaskus [11]

Answer:

because it’s different system

Step-by-step explanation: When going from the metric system to another, You do conversion factor.

8 0
3 years ago
Read 2 more answers
Find the slope of the line
Reika [66]

Answer:

7/3

Step-by-step explanation:

We can find the slope of a line given two points by

m = (y2-y1)/ (x2-x1)

    = (3--4)/(4-1)

   = (3+4)/(4-1)

7/3

5 0
3 years ago
Read 2 more answers
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
What is the value of h?
emmainna [20.7K]
The angle on one side of a straight line is 180°, so 40° + 60° + h = 180°.
If you rearrange this you will get h = 180° - 60° - 40°, this will give you: h = 80°.
Hope this helps
5 0
2 years ago
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