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2 years ago

What is the unit rate? Will give brainliest

Mathematics

1 answer:

givi [52]2 years ago

6 0

“Unit rate” is a comparison of any two separate but related measurements when the second of these measurements is reduced to a value of one. Calculating the unit rate in any set of circumstances will require the use of division.

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If f(x) = x-1/3 and g(x)= 3x+1, what is (f o g)(x)?

N76 [4]

Answer:

$(f o g)(x) = 3x + \frac{2}{3}$

Step-by-step explanation:

We have the function $f(x) = x-\frac{1}{3}$ and we have the function $g(x) = 3x + 1$ . We want to find g(x) composed with f(x)

Then, the function (f o g)(x) is the same since f(g(x))

That is, you must do x = g(x) and then enter g(x) into the function f(x).

$f(g(x)) = (g(x)) -\frac{1}{3}$

$f(g(x)) = (3x + 1) -\frac{1}{3}$

Simplifying, we obtain:

$(f o g)(x) = 3x + 1 -\frac{1}{3}\\\\(f o g)(x) = 3x + \frac{2}{3}$

Finally. The composite function is:

$(f o g)(x) = 3x + \frac{2}{3}$

6 0

2 years ago

How would i solve. been getting wrong

Lisa [10]

Answer:just find a common denomintor and add them

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

The coordinates of the vertices of a regular polygon are given. Find the area of the polygon to the nearest tenth.

alexandr1967 [171]

Answer:

The area is equal to $8\ units^{2}$

Step-by-step explanation:

we have

A(0, 0), B(2, -2), C(0, -4), D(-2, -2)

Plot the figure

The figure is a square (remember that a regular polygon has equal sides and equal internal angles)

see the attached figure

The area of the square is

$A=AB^{2}$

Find the distance AB

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

substitute the values

$AB=\sqrt{(-2-0)^{2}+(2-0)^{2}}$

$AB=\sqrt{(-2)^{2}+(2)^{2}}$

$AB=\sqrt{8}$

$AB=2\sqrt{2}\ units$

Find the area of the square

$A=(2\sqrt{2})^{2}$

$A=8\ units^{2}$

4 0

2 years ago

Find all the zeros if the function including multiplicity f(x)=(x+7)(x-3)^2(x-2+i)(x-2-i)

NISA [10]

Solve for x. The multiplicity of a root is the number of times the root appears.
x=-7 (multiplicity of 1)
x=3 (multiplicity of 2)
x=2-i (multiplicity of 1)
x=2+1 (multiplicity of 1)

6 0

2 years ago

Consider the implicit differential equation <img src="https://tex.z-dn.net/?f=%2849%20y%5E%7B3%7D%20%2B%2045%20xy%29%20dx%20%2B%

BaLLatris [955]

We're looking for an integrating factor $\mu(x,y)=x^py^q$ such that

$\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0$

is exact, which would require that

$(\mu M)_y=(\mu N)_x$
$(49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x$
$49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q$
$\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4$

You can verify that $(\mu M)_y=(\mu N)_x$ if you'd like. With the ODE now exact, we have a solution $F(x,y)=C$ such that

$F_x=\mu M$
$F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx$
$F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)$

$F_y=\mu N$
$50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4$
$f'(y)=0$
$\implies f(y)=C$

and so the general solution is

$F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C$

8 0

2 years ago