Show all ways to make 52 as tens and ones

1 answer:

4 0

There are a very large amount of possible solutions. A few examples.
10 + 10 + 10 + 10 + 10 + 1 + 1 = 52.
10 + 10 + 10 + 10 + (12 ones here) = 52.
10 + 10 + 10 + (22 ones here) = 52.
10 + 10 + (32 ones here) = 52
10 + (42 ones here) = 52
(52 ones here) = 52

I'm fairly sure those are all possible ways.

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There is a bag with only milk and dark chocolates. The probability of randomly choosing a dark chocolate is 5 12 . There are 40

Maksim231197 [3]

Answer:

There must be 96 milk chocolates.

Step-by-step explanation:

If you take the original ratio, 5:12, and replace the 5 with 40, you'd discover that 40 divided by 5 is 8. So what do you do? You multiply the 12 by 8 to get 96 so now the ratio is 40:96.

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.

_____

Additional comment

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.