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solniwko [45]
3 years ago
7

The equation y = −2x + 3 is the boundary line for the inequality y ≤ −2x + 3. Which sentence describes the graph of the inequali

ty?
A. The region shaded above a dashed boundary line.


B. The region shaded above a solid boundary line.


C. The region shaded below a solid boundary line.


D. The region shaded below a dashed boundary line.

Mathematics
2 answers:
bulgar [2K]3 years ago
8 0

Answer:

C. The region shaded below a solid boundary line.

Step-by-step explanation:

The given inequality is

y\le-2x+3

The boundary line for this inequality is y=-2x+3.

Since the inequality involve \le, we use a solid boundary line.

After graphing the boundary line; we test the origin to determine which half -plane to be shaded. This is because the boundary lie is not passing through the origin.

Testing the origin yields 0\le-2(0)+3.

This implies that;  0\le3. This statement is true.

Hence we shade the lower half-plane of the solid boundary line.

The correct choice is C.

See graph

lukranit [14]3 years ago
6 0

Answer: Option C

Step-by-step explanation:

The point-slope form of the equation of the line is:

y=mx+b

Where m is the slope and b is the intersection with the y-axis.

In the line y=-2x + 3, you can identify that:

m=-2\\b=3

The symbol of the inequality "\leq" indicates that you must shade the region below the boundary line and for the symbols of inequalities "\leq" and "\geq" the line must be solid (Observe the graph attached).

Then the answer is the option C.

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How does replacing f(x) with f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative) affect
ivann1987 [24]
F(x) + k - Moves the graph k units up.

k f(x)   stretches the graph parallel to y-axis by a facor k

f (kx) stretches  the  graph by a factor 1/k parallel to x-axis

f(x + k)  moves the graph   3 units to the left.

For k negative the first one moves it k units down

for second transform negative does same transfoormation but also reflects the graph in the x axis

For the third transform negative k :- same as above but also reflects in y axis

4th transform -  negative k moves graph k units to the right


4 0
3 years ago
If the initial amount of iodine-131 is 537 grams , how much is left after 10 days?
viktelen [127]

Answer:

225.78 grams

Step-by-step explanation:

To solve this question, we would be using the formula

P(t) = Po × 2^t/n

Where P(t) = Remaining amount after r hours

Po = Initial amount

t = Time

In the question,

Where P(t) = Remaining amount after r hours = unknown

Po = Initial amount = 537

t = Time = 10 days

P(t) = 537 × 2^(10/)

P(t) = 225.78 grams

Therefore, the amount of iodine-131 left after 10 days = 225.78 grams

4 0
3 years ago
Express the ratio 4:7 in the form n:1
Snowcat [4.5K]

Answer:

11:1

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Factor the expression. <br><br> 9x^2 + 48x + 64
Savatey [412]
(3x+8)(3x+8) is the answer :))
4 0
3 years ago
PLEASE HELP!!<br> Write Y=x^2+4x+3 in standard form
DanielleElmas [232]

Answer:

already in standard form

Step-by-step explanation:

the equation of a quadratic in standard form is

y = ax² + bx + c ( a ≠ 0 )

y = x² + 4x + 3 ← is in standard form

with a = 1 , b = 4 , c = 3

6 0
2 years ago
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