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Bingel [31]
3 years ago
11

Someone please help me I’m actually struggling so much rn.

Mathematics
1 answer:
alexandr1967 [171]3 years ago
7 0

Answer:

Step-by-step explanation:

okay don't feel bad, you actually got it right, let me help you with some holes!

You calculate slope by the rise over run. These people are confusing you a bit - do problem 2 then 1

the vertical change is -1 horizontal change is 3 so by rise/run=-1/3

-1/3 is the slope

for the second problem this is confusing but just read and try to analyze

vertical change/horizontal change = 5-4/0-3= 1/-3(or -1/3)

The y intercept is the point where the line hits the y axis, so at point (0, 5)

The function is y=mx+b

m is the slope

b is the y intercept

so y=-1/3x+5

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On a number line, graph all the temperature that are warmer than -2 degrees Celsius
Bogdan [553]
The inequality would be x>2
5 0
3 years ago
Help pleaseee ASAP ASAP
Yanka [14]

Answer:

I think the answer is the third one

Step-by-step explanation:

7 0
3 years ago
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
Can someone help me on this one pls
frozen [14]

Answer:

d=5

Step-by-step explanation:

6x = 5

d = 6x

if 6x = 5 then d would also equal 5

3 0
3 years ago
Read 2 more answers
6^2+17+12 help me plz.
Dmitry_Shevchenko [17]
6^2 + 17 + 12

First, let's start with 6^{2} . Basically, the '2' above the 6 indicates that the number appears twice in multiplying. 6^{2} can be described as 6 × 6 as well. 36 should be your answer for 6 squared.
6^{2} = 6 \times 6 = 36

Second, our problem should look like: 36 + 17 + 12. One can simply do this on a calculator or count on their fingers. Once you add all of the numbers up, you should get 65.

Answer: \fbox {65}
4 0
3 years ago
Read 2 more answers
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