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stepan [7]
3 years ago
13

Write an inequality for the graph.

Mathematics
2 answers:
Rainbow [258]3 years ago
8 0

Answer:

y\geq |x-5|+1

Step-by-step explanation:

From the graph it is clear that it is a V-shaped curve. So, the given graph represents an absolute value function.

The vertex form of an absolute function is

y=|x-h|+k

where, (h,k) is vertex.

The vertex of the given graph is (5,1). So the related equation of the graph is

y=|x-5|+1

Related line is a solid line and shaded region lie above the line, So the sign of inequality must be ≥.

y\geq |x-5|+1

Check by (0,0).

0\geq |0-5|+1

0\geq 6

It is a false statement.

Therefore, y\geq |x-5|+1.

icang [17]3 years ago
7 0

Answer:

The answer to your question is:

The inqualites are       y > -x + 6    and    y > x - 4

Step-by-step explanation:

Data

Look for the equations of the lines

First line

A (5, 1)

B (0,6)

m = (6 - 1) /(0 - 5)

m = 5 / -5

m = -1

 y- y1 = m(x - x1)

y - 1 = -1(x - 5)

y -1 = -x + 5

y = -x + 5 + 1

 y = -x + 6    The shadow are is upward the line then

y > -x + 6

Second line

A (5, 1)

B (6, 2)

m = (2 - 1) / (6 - 5)

m = 1 / 1 = 1

y - 1 = 1(x - 5)

y = x - 5 + 1

y = x - 4   The shadow must be upward the line, then

y > x - 4

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23.55 + 19.95 = 43.1 (total amount spent)

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50 - 43.1 = 6.9

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compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.
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\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})

We have been given two vectors $\vec{a}$ and $\vec{b}$, we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of$\vec{b}$onto $\vec{a}$means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

#SPJ4

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