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lidiya [134]
4 years ago
6

What type of transformation transforms (a,

Mathematics
1 answer:
alisha [4.7K]4 years ago
5 0
The answer would be reflection over the y-axis because, the x changes while the y stays the same
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G find the domain for the particular solution to the differential equation dy dx equals the quotient of negative 1 times x and y
faust18 [17]
\frac{dy}{dx} = \frac{-x}{y}
by separating the variables
∴ y dy = -x dx
integrating both sides with respect to x
∴ ∫y dy = ∫-x dx
∴ \frac{ y^{2} }{2} = - \frac{ x^{2} }{2} +c
finding c using initial condition y(2) = 2.
∴ \frac{ 2^{2} }{2} = - \frac{ 2^{2} }{2} +c
∴ c = 4
 ∴ \frac{ y^{2} }{2} = - \frac{ x^{2} }{2} +4
multiplying all the equation by 2
∴ y² = -x² + 8
∴ x² + y² = 8
The resultant equation represents the equation of the circle
with center (0,0)  and radius = √8 = 2√2
so, the domain of this function is [-2√2 , 2√2]
See the attached figure



7 0
3 years ago
Determine whether the graph represents a function . If it does represent a function , give its domain and range.
laiz [17]

Answer:

A function is a relationship that maps elements from a set (the domain) into elements from another set (the range)

Such that each element in the domain can be mapped into only one element from the range.

Let's see graphs 18, 20,24 and 30.

Remember that the axis that represents the domain is the horizontal one (usually represented with x), and the vertical axis represents the range (usually represented with y)

18) Here we can see that the point x = 1 his mapped into two different values of y.

we have the pair (1, 2) and the pair (1, -3)

And something similar happens for x = 2.

Then we can conclude that this is not a function.

20) Here we have a linear relationship.

Linear relationships are almost always functions, the only case when these are not functions is when the linear equation is something like x = a.

Linear equations can be written as:

y = a*x + b

So x can be any value, and thus y also can be any value.

Then the domain is the set of all real numbers, and the range is the set of all real numbers.

24) Here we have a quadratic function whose arms go down.

This is a function, now let's see the domain and range.

Quadratic functions are written as:

y = a*x^2 + b*x + c

There is no value of x can cause some problem in this equation, then this function works for all values of x, then the domain is the set of all real numbers.

Now, let's look at the graph.

We can see that the function goes up, reaches a maximum, and then goes down again.

Then the range will be the set of all the values smaller than the maximum we can see in the graph, this is:

y ∈ (-∞, 15]

or simply:

y ≤ 15.

30) Here again, we can see that for x = 0 there are two different values of y.

the same happens for x = -1, x = -2, and a lot of other values.

Then this is not a function, because it is mapping values of the domain into different values of the range.

3 0
3 years ago
If X =4 in., find the area of the trapezoid in problem 22
Digiron [165]

Answer:

Need a picture of the problem.

Step-by-step explanation:

5 0
3 years ago
Which test point holds true for y − 2x ≤ 1?
Nataly_w [17]
I think it is (-2,4)
8 0
4 years ago
4. Simplify the following.<br>3<br>a. 2-X5-:11<br>3<br>x5<br>5<br>6<br>7​
Ulleksa [173]

Answer:

<h2>1 \frac{1}{4}</h2>

Step-by-step explanation:

2 \frac{3}{7}  \times  5\frac{5}{6}   \div 11 \frac{1}{3}

Convert the mixed number to an improper fraction

\frac{17}{7}  \times  \frac{35}{6}  \div  \frac{34}{3}

To divide by a fraction, multiply the reciprocal of that fraction

\frac{17}{7} \times  \frac{35}{6}  \times  \frac{3}{34}

Reduce the number with the G.C.F 7

17 \times  \frac{5}{6}  \times  \frac{3}{34}

Reduce the numbers with the G.C.F 17

\frac{5}{6}  \times  \frac{3}{2}

Reduce the numbers with the G.C.F 3

\frac{5}{2}  \times  \frac{1}{2}

Multiply the fraction

\frac{5}{4}

In mixed fraction:

1 \frac{1}{4}

Hope this helps..

Good luck on your assignment...

4 0
3 years ago
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