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2 years ago

I need help with this

Mathematics

1 answer:

andrey2020 [161]2 years ago

3 0

18a:16 18b:32 % means out of 100

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A 6 pound bag of oranges cost 6.78 What is the unit price?

oksano4ka [1.4K]

Answer:

$1.13

Step-by-step explanation:

Since 6 pound bag of oranges costs $6.78...

$\frac{6.78}{6}$

Now, let's divide 6.78 in to 6 to find our unit price.

$\frac{1.13}{1}$

Now we know 1 pound, (our unit price), costs...

$$1.13$

5 0

3 years ago

What is the correct interpretation of the point (250, 0)? The average price per ticket is $250. The minimum price per ticket is

Brut [27]

The point (250,0) of the graph represents that the average price per ticket is $250.

Given to us

x is the price the passenger paid

f(x) is the positive percent difference

<h3>What is the correct interpretation of the point (250, 0)?</h3>

We know that a coordinate is written in the form of (x, y), therefore, the point (250, 0) represents that the price of the ticket is 250, while the 0 in the coordinate represents that there is no percentage difference. Since the point (250,0) is the mid-value of the x-axis on the graph, we can say that $250 is the average price of the ticket.

Hence, the point (250,0) of the graph represents that the average price per ticket is $250.

Learn more about Graph:

brainly.com/question/14375099

5 0

2 years ago

An initial population of 535 quail increases at an annual rate of 25%. Write an exponential function to model the quail populati

olchik [2.2K]

Answer: c

Step-by-step explanation:

5 0

2 years ago

Could you please help me for this question?

Olin [163]

Answer:

See attached for graphs

g(x) -- domain: -∞ < x < ∞; range: 0 < y < ∞

g^-1(x) -- domain: 0 < x < ∞; range: -∞ < y < ∞

Step-by-step explanation:

g(x) is an exponential decay function. Its base is 1/3, so each increase of 1 unit in x will multiply the y-value by a factor of 1/3. The graph will rapidly approach its horizontal asymptote of y=0 as x gets large. The y-intercept is (0, 1). Just as y gets smaller as x increases, so it gets larger as x decreases. Each decrease of x by 1 unit causes the y-value to be multiplied by 3.

The graph of g^-1(x) is the graph of g(x) reflected across the line y=x. That is, each coordinate pair (x, y) on the graph of g(x) becomes a point (y, x) on the graph of the inverse function. In order to graph g^-1(x), you don't need to write down the function, you only need to know the relationship between the graphs.

Just as x- and y- are interchanged on the graph, so the domain, range, and intercepts are interchanged. g^-1(x) will have a vertical asymptote of x=0, and an x-intercept of (1, 0). The domain of g^-1(x) is the range of g(x): 0 < x < ∞; and the range of g^-1(x) is the domain of g(x): -∞ < y < ∞.

The attached graph shows g(x) in red and g^-1(x) in blue. As you can see, we created the graph simply by interchanging x and y. The line y=x is shown for reference, so you can see that each curve is a reflection of the other across that line.

_____

<em>Additional comment</em>

The explicit expression for g^-1(x) can be found by solving for y:

x = g(y)

$x=\left(\dfrac{1}{3}\right)^y=\dfrac{1}{3^y}=3^{-y}\\\\ \log(x)=-y\cdot\log(3)\qquad\text{take logarithms}\\\\y=-\dfrac{\log{x}}{\log{3}}=-\log_3{x}\qquad\text{use the change of base relation}\\\\\boxed{g^{-1}(x)=-\log_3{x}}$

If you're familiar with the log function, you know it has an x-intercept of 1 and a vertical asymptote at x=0. The base of the log function is simply a vertical scale factor. The minus sign reflects it across the x-axis.

6 0

2 years ago

If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(

timama [110]

Use the change-of-basis identity,

$\log_x(y) = \dfrac{\ln(y)}{\ln(x)}$

to write

$xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}$

Use the product-to-sum identity,

$\log_x(yz) = \log_x(y) + \log_x(z)$

to write

$xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}$

Redistribute the factors on the left side as

$xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}$

and simplify to

$xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)$

Now expand the right side:

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}$

Simplify and rewrite using the logarithm properties mentioned earlier.

$xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1$

$xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}$

$xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}$

$xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)$

$\implies \boxed{xyz = x + y + z + 2}$

(C)

6 0

1 year ago