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2 years ago

No it wouldnt because quadrant one is ++ so how is it that

Mathematics

2 answers:

Whitepunk [10]2 years ago

5 0

Answer: no

This is nothing but a meretricious deceit, for my solemn is nothing but fathomable and ostentatious. This is true to the extend of ontologism and Kiergegaardism, thereby through all of certainty. You shall be of verboten and censured into utter dissension, for your imbroglio is nothing but a unilateral anathema, antithesis; recherche. This is to the periphery of abomination, to the brink of Existentialism and ethical egoism. For thou must be interdicted in all of certainty. Though it may seem a concomitant, let be a contumacious individual. This shall not avail, such defamatory and garrulous desultory shall not prevail. Thus, it shall be obliterated fecklessly with no lachrymose, for it shall never be remembered.

nexus9112 [7]2 years ago

4 0

Answer:

what do you mean mate? I don't quite get it sorry.

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Find the distance from A to C

lora16 [44]

Answer:

Step-by-step explanation:

4 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 are the coordinates of the turning point for the function f(x) = (x + 2)3 − 4?

AnnZ [28]

Answer:A

Step-by-step explanation:

Given

$F(x)=(x+2)^3-4$

Turning Point is the point where the graph changes its nature i.e. either increasing to decreasing or decreasing to increasing

to find turning point

$F'(x)=3(x+2)^2+0$

For critical point

F'(x)=0

$3(x+2)^2=0$

x+2=0

x=-2

$F(x)=(-2+2)^3-4=-4$

thus

(-2,-4 ) is the turning Point where Function changes its nature

7 0

3 years ago

a student Miss 10 problems on an English test and received a grade of 44%. If all the problems were of equal value, how many pro

eimsori [14]

Answer:

Rounded to the nearest integer, the test had 18 problems in all.

Step-by-step explanation:

Given that a student missed 10 problems on an English test and received a grade of 44%, if all the problems were of equal value, to determine how many problems were on the test the following calculation must be performed:

100 - 44 = 56

56 = 10

100 = X

100 x 10/56 = X

1,000 / 56 = X

17.85 = X

Thus, rounded to the nearest integer, the test had 18 problems in all.

5 0

3 years ago

Helllllppppppppppppp

antoniya [11.8K]

Answer:

0.62

Step-by-step explanation:

0.5+0.1+.02=0.62

5 0

3 years ago