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3 years ago

What is the domain and range ?

Mathematics

1 answer:

Svetlanka [38]3 years ago

3 0

Answer:

Domain is [-4, 4]

Range is [0, 4]

Step-by-step explanation:

Domain is the fancy name for the values of x that works for the graph (input), in this case, from -4 to 4, inclusive.

Range is the fancy name for the values of y that works for the graph (output), in this case, from 0 to 4, inclusive.

You might be interested in

In right triangle RQS, ∠Q is a right angle, RQ=55, QS=48, and RS=73.

kari74 [83]

The ratio of cos S in triangle RQS is 48 / 73

<h3 /><h3>Right angle triangle:</h3>

It has one angle as 90 degree
The sides can be found using Pythagoras theorem

∠Q = 90 degrees

Therefore, Finding cos S,

opposite side = RQ = 55

Hypotenuse = RS = 73

Adjacent side = QS = 48

Therefore,

cos S = adjacent / hypotenuse

cos S = 48 / 73

learn more on right angle triangle: brainly.com/question/13907707?referrer=searchResults

5 0

2 years ago

How do you do (a) and (b)?

bulgar [2K]

Answer:

See solution below

Step-by-step explanation:

(a) If n=0 or 1, the equation

(1) $y' = a(t)y + f(t)y^n$

would be a simple linear differential equation. So, we can assume that n is different to 0 or 1.

Let's use the following substitution:

(2) $z=y^{n-1}$

Taking the derivative implicitly and using the chain rule:

(3) $z'=(1-n)y^{-n}y'$

Multiplying equation (1) on both sides by

$(1-n)y^{-n}$

we obtain the equation

$(1-n)y^{-n}y' = (1-n)y^{-n}a(t)y+(1-n)y^{-n}f(t)y^n$

reordering:

$(1-n)y^{-n}y' = (1-n)y^{-n}ya(t)+(1-n)y^{-n}y^nf(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)y^{0}f(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)f(t)$

Now, using (2) and (3) we get:

$z'= (1-n)za(t)+(1-n)f(t)$

which is an ordinary linear differential equation with unknown function z(t).

(b)

The equation we want to solve is

(4) $xy'+ y = x^4 y^3$

Here, our independent variable is x (instead of t)

Assuming x different to 0, we divide both sides by x to obtain:

$y'+\frac{1}{x}y = x^3 y^3$

$y' = -\frac{1}{x}y+x^3 y^3$

Which is an equation of the form (1) with

$a(x)=-\frac{1}{x}$

$f(x)=x^3$

$n=3$

So, if we substitute

$z=y^{-2}$

we transform equation (4) in the lineal equation

(5) $z'=\frac{2}{x}z-2x^3$

and this is an ordinary lineal differential equation of first order whose

integrating factor is

$e^{\int (-\frac{2}{x})dx}$

but

$e^{\int (-\frac{2}{x})dx}=e^{-2\int \frac{dx}{x}}=e^{-2ln(x)}=e^{ln(x^{-2})}=x^{-2}=\frac{1}{x^2}$

Similarly,

$e^{\int (\frac{2}{x})dx}=x^2$

and the general solution of (5) is then

$z(x)=x^2\int (\frac{-2x^3}{x^2})dx+Cx^2=-2x^2\int xdx+Cx^2=\\\ -2x^2\frac{x^2}{2}+Cx^2=-x^4+Cx^2$

where C is any real constant

Reversing the substitution

$z=y^{-2}$

we obtain the general solution of (4)

$y=\sqrt{\frac{1}{z}}=\sqrt{\frac{1}{-x^4+Cx^2}}$

Attached there is a sketch of several particular solutions corresponding to C=1,4,6

It is worth noticing that the solutions are not defined on x=0 and for C<0

4 0

3 years ago

The ratio x : y is equal to 2 : 5, and y is 80% of z. Write an equation to represent the proportional relationship between x and

Mila [183]

Answer:

x : z = 8 : 25

Step-by-step explanation:

<u>Given</u>

x : y = 2 : 5, y is 80% of z

<u>Rearranging the ratios:</u>

x/y = 2/5 ⇒ x/y = 0.4 ⇒ y = x/0.4 ⇒ y = 2.5x
y is 80% of z ⇒ y = 0.8z

<u>Comparing the values of y:</u>

2.5x = 0.8z
x/z = 0.8/2.5
x/z = 0.32
x/z = 32/100
x/z = 8/25
x : z = 8 : 25

4 0

3 years ago

A bracelet is made by stringing together four beads. Each bead is either red or green. How many different color patterns are

JulijaS [17]

Answer:

3 or 5, depends on the bracelet can be fully green or red. If you have to have both colours then it's 3.

Step-by-step explanation:

RRRG

RRGG

RGGG

GGGG?

RRRR?

7 0

3 years ago

Read 2 more answers

B) y = x2 - 10<br> d) y = x3 - 9<br> f) y = (x - 2)2<br> h) y = (2x + 3)

nika2105 [10]

What exactly are you asking? Can you maybe describe it more please?

6 0

3 years ago