How do I graph y is less than -1 on a number graph

Mathematics

1 answer:

lana66690 [7]3 years ago

6 0

I used a dash line because the term 'less than' does not include the value itself.

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Please help me ASAPP

swat32

Answer:

I believe the answer is, A.

Explanation:

Ordered pairs are classified as a function when none of the x’s in the ordered pairs match.

Like in b, a pair has 3.
And in c, two pairs have 1 and 3.

7 0

2 years ago

1. Which ordered pairs are solutions to the inequality –3x + y ≥ 7?

Murrr4er [49]

Q1. The answers are (–1, 8), (0, 7), (3, 18)

–3x + y ≥ 7
Let's go through all choices:

(–2, –3)
(-3) * (-2) + (-3) ≥ 7
6 - 3 ≥ 7
3 ≥ 7 INCORRECT

(–1, 8)
(-3) * (-1) + 8 ≥ 7
3 + 8 ≥ 7
11 ≥ 7 CORRECT

(0, 7)
(-3) * 0 + 7 ≥ 7
0 + 7 ≥ 7
7 ≥ 7 CORRECT

(1, 9)
(-3) * 1 + 9 ≥ 7
-3 + 9 ≥ 7
6 ≥ 7 INCORRECT

(3, 18)
(-3) * 3 + 18 ≥ 7
-9 + 18 ≥ 7
9 ≥ 7 CORRECT

Q2. The answers are:
5x + 12y ≤ 80
x ≥ 4
y ≥ 0

x - small boxes
y - large boxes

He has x small boxes that weigh 5 lb each and y large boxes that weigh 12 lb each on a shelf that holds up to 80 lb:
5x + 12y ≤ 80

Jude needs at least 4 small boxes on the shelf: x ≥ 4

Let's check if y can be 0:
5x + 12y ≤ 80
5x + 12 * 0 ≤ 80
5x + 0 ≤ 80
5x ≤ 80
x ≤ 80 / 5
x ≤ 16

x ≥ 4 can include x ≤ 16

So, y can be 0: y ≥ 0

8 0

2 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):