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3 years ago

The curve with equation y = f(x) is reflected and squashed so that the point (1,3)

Mathematics

1 answer:

Natali5045456 [20]3 years ago

4 0

Answer:

I don't know

Step-by-step explanation:

I am 8 class student sorry

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A plumber charges $50 to make a house call. He also charges $25 per hour for labor. Write an equation that represents x, the num

kari74 [83]

t = total

equation: t = 25x + 50

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4 years ago

Emily designed a robot that could climb down steps. She drew a graph to show the results from a test of her new robot. The graph

Artist 52 [7]

The generic equation of the line is:

$y = mx + b$

Where,

m: slope of the line.

b: cutting point with vertical axis.

The slope of the line is given by:

$m =\frac{y2-y1}{x2-x1}$

$m =\frac{14-15}{10-0}$

$m =-\frac{1}{10}$

The cutting point with the vertical axis is:

$b=15$

Therefore, the equation of the line is:

$y =-\frac{1}{10}x + 15$

Answer:

the equation that represents Emily’s situation is:

$y =-\frac{1}{10}x + 15$

5 0

3 years ago

Read 2 more answers

Find the closest point to y in the subspace W spanned by v1 and v

enot [183]

Answer:

The answer is

$\bold{ \left[\begin{array}{c}\frac{127}{27} \\ \ &\frac{65}{18} \\ \ &\frac{76}{27}\\ \ &\frac{97}{54}\\\ \end{array}\right]}$

Step-by-step explanation:

Given value:

$v_1=\left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \\$

$v_2=\left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] \\$

The value of $v_1 \cdot v_2$ :

$= \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]$

The above value " $v_1, v_2$ " are orthogonal vectors that is: $v_1 \neq 0, \ \ v_2 \neq 0$

$\{v_1,v_2\}$ from the above orthogonal basis of subspace w and the shortest distanc value of vector y:

$=\left[\begin{array}{c}3&9&-5&8\\\ \end{array}\right] \\\\$

The value of y on $W= \frac{y \cdot v_1}{v_1 \cdot v_1} \times v1 + \frac{y \cdot v_2}{v_2 \cdot v_2} \times v_2$

$= \frac{\left[\begin{array}{c}3&9&-5&8\\\ \end{array}\right] \cdot \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] }{\left[\begin{array}{c}5&4&3&2\\ \end{array}\right] \cdot \left[\begin{array}{c}5&4&3&2\\ \end{array}\right]} \times \left[\begin{array}{c}5&4&3&2\\ \end{array}\right] +$ $\frac{\left[\begin{array}{c}3&9&5&-8\\\ \end{array}\right] \cdot \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] }{\left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right] \cdot \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]} \times \left[\begin{array}{c}-4&5&-2&3\\ \end{array}\right]$