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4 years ago

What is used when graphing an inequality that uses one of these two symbols: ≤, ≥ ?

1 answer:

3 0

Since <span>using ≤, ≥</span>, it means that the lines are included, the line is graphed as a solid line. When you have plain > or <, the line is not included, so you use a dashed line.

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Easy Points<br><br> whats 1=1

Katen [24]

infinite set is the answer

6 0

3 years ago

Read 2 more answers

jenny went to the mall with $120 in her wallet.she spent $25.65 on a shirt, $7.98 on a pair of socks, $16.23 on a cd , and $4.58

Effectus [21]

Total money=$120
on shirt=$25.65
on socks=$7.98
on a cd=$16.23
on snack=$4.58
so total money left=total money-total money spent
=$120-($25.65+$7.98+$16.23+$4.58)
=$120-$54.44
so total money left=$65.56

4 0

3 years ago

Read 2 more answers

How many solutions does 6x−1=34−x have

aleksley [76]

Step-by-step explanation:

6x+x=34+1

7x=35

x=35/7

x=5

Hence this equation has only one solution

4 0

3 years ago

What is the range of the function represented by this graph?

Lera25 [3.4K]

Answer:

A. y ≤ 1

Step-by-step explanation:

according to the graph, the coordinates of maximum value is (-3, 1) and besides the graph towards down, so, the range of the function is y ≤ 1

4 0

3 years ago

HELP!! Algebra help!! Will give stars thank u so much <333

Anna35 [415]

Answers:

Part a) $\bf{\sqrt{x^2+(x^2-3)^2}$
Part b) 3
Part c) 2.24
Part d) 1.58

============================================================

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

$d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\$

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying $d = \sqrt{x^2 + (x^2-3)^2}$ is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\$

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

$d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\$

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0

3 years ago