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goblinko [34]
2 years ago
12

Someone help me please:)

Mathematics
1 answer:
Svetradugi [14.3K]2 years ago
5 0

Answer:

your answer should be C

Step-by-step explanation:

its gonna be in negative

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Which graph represents the solution set of the system of inequalities? {x+y<12 y≥x−4
RSB [31]

Answer: The graph in the bottom right-hand corner

(see figure 4 in the attached images below)

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

Explanation:

Let's start off by graphing x+y < 1. The boundary equation is x+y = 1 since we simply change the inequality sign to an equal sign. Solve for y to get x+y = 1 turning into y = -x+1. This line goes through (0,1) and (1,0). The boundary line is a dashed line due to the fact that there is no "or equal to" in the original inequality sign. So x+y < 1 turns into y < -x+1 and we shade below the dashed line. The "less than" means "shade below" when y is fully isolated like this. See figure 1 in the attached images below.

Let's graph 2y >= x-4. Start off by dividing everything by 2 to get y >= (1/2)x-2. The boundary line is y = (1/2)x-2 which goes through the two points (0,-2) and (4,0). The boundary line is solid. We shade above the boundary line. Check out figure 2 in the attached images below.

After we graph each individual inequality, we then combine the two regions on one graph. See figure 3 below. The red and blue shaded areas in figure 3 overlap to get the purple shaded area you see in figure 4, which is the final answer. Any point in this purple region will satisfy both inequalities at the same time. The solution point cannot be on the dashed line but it can be on the solid line as long as the solid line is bordering the shaded purple region. Figure 4 matches up perfectly with the bottom right corner in your answer choices.

5 0
2 years ago
Read 2 more answers
1) State the amplitude, period, phase shift, vertical shift and graph
bogdanovich [222]

Answer:

See answers and explanations below (along with a visual graph)

Step-by-step explanation:

For the function y=acos(b(x+c))+d:

Amplitude: |a|

Period: \frac{2\pi}{|b|}

Phase shift: \frac{-c}{b}

Vertical shift: d

Therefore, for y=2cos(\frac{2}{3}(\theta+\pi))-1:

Amplitude: |a|=|2|=2, which is 2 units above and below the midline (see d)

Period: \frac{2\pi}{|b|}=\frac{2\pi}{|\frac{2}{3}|}=\frac{2\pi}{\frac{2}{3}}=3\pi, so the cycle will repeat every 3π units

Phase shift: \frac{-c}{b}=\frac{-\frac{2\pi}{3} }{\frac{2}{3} }=-\pi, or π units to the left

Vertical shift: d=-1, or 1 unit down

3 0
2 years ago
Graph the system &amp; write its solution <br> 2x+y=-4<br> Y=-1/2x-1
xz_007 [3.2K]

Answer:

The solution of system of equation is (-2,0)

Step-by-step explanation:

Given system of equation are

Equation 1 :      2x+y=(-4)

Equation 2 :      y+\frac{1}{2}x=(-1)

To plot the equation of line, we need at least two points

For Equation 1 : 2x+y=(-4)

Let x=0

2x+y=(-4)

2(0)+y=(-4)

y=(-4)

Let x=1

2x+y=(-4)

2(1)+y=(-4)

y=(-6)

Therefore,

The required points for equation is (0,-4) and (1,-6)

For Equation 2 : y+\frac{1}{2}x=(-1)

Let x=0

y+\frac{1}{2}x=(-1)

y+\frac{1}{2}(0)=(-1)

y=(-1)

Let x=2

y+\frac{1}{2}x=(-1)

y+\frac{1}{2}(2)=(-1)

y=(-2)

The required points for equation is (0,-1) and (2,-2)

Now, plot the graph using this points

From the graph,

The red line is equation 1 and blue line is equation 2

Since. The point of intersection is solution of system of equations

The solution of system of equation is (-2,0)

6 0
3 years ago
What is the axis of symmetry of g(x)=2(x-4)(x+2)
maksim [4K]
I think the answe would be 4(56x)
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Identify the hypothesis in the statement If it is monday, then the basketball team is playing
dangina [55]
“Then the basket ball team is playing” is the hypothesis because that is the guess being made
6 0
3 years ago
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