Ask question

Ask question

All categories

3 years ago

8

Graph the solution set of this inequality: 5x − 4y > 20.

1 answer:

Yanka [14]3 years ago

3 0

Answer:

Step-by-step explanation:

Solve the inequality 5x − 4y > 20 for y, as follows: Subtract 5x from both sides, obtaining:

-4y > 20 - 5x;

Then divide all terms by -4:

y < -5 +(5/4)x, where the direction of the inequality sign has been reversed because of division by a negative quantity.

Temporarily replace the < symbol with = obtaining y = -5 +(5/4)x. Now choose at least three x values and find the corresponding y values. For example:

x y = -5 +(5/4)x

0 -5

4 0

-8 -15

Now plot these three points (0, -5), (4, 0) and (-8, -15). Draw a dashed line through them. Because of the < symbol in y < -5 +(5/4)x, shade the area underneath the dashed line.

You might be interested in

Which figures can have the same cross section as a sphere?

shepuryov [24]

Answer:

2

Step-by-step explanation:

the cross section of a sphere is a circle

only the cone can have a circle for a cross section

8 0

3 years ago

PLSSS help quickly worth 20 pts

gayaneshka [121]

1.09 that's the answer ! I just wrote extra stuff for the sake of it . x= 1.09

7 0

2 years ago

The height (h), in feet, of a balloon is calculated by using the equation h equals 3t + 2 where t is the time,in seconds. What i

navik [9.2K]

ANSWER: 92

EXPLANATION:
You plug in 30 where t us and solve.
3(30)+2=92

4 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

1 year ago

Between which two numbers is Negative two-thirds located on a number line?

klemol [59]

Answer:-1 and 0

Step-by-step explanation:

5 0

3 years ago