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Luba_88 [7]
3 years ago
9

Evaluate -4a + 10 when a = -2

Mathematics
1 answer:
stiks02 [169]3 years ago
4 0

Answer:

<u>18</u>

Step-by-step explanation:

<em>replace a with -2</em>

-4(-2) + 10

<em>multiply the -4 and the -2, the negatives cancel out and leave you with 8</em>

8 +10

<em>just add</em>

18

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Find the equation of the line that is perpendicular to x - y = -5 and passes through (4, 7).
Ivanshal [37]
X - y = -5
-y = -x - 5
y = x + 5...slope here is 1. A perpendicular line will have a negative reciprocal slope. All that means is flip ur slope and change the sign. So our perpendicular line will have a slope of -1.

y = mx + b
slope(m) = -1
(4,7)...x = 4 and y = 7
now we sub and find b, the y int
7 = -1(4) + b
7 = -4 + b
7 + 4 = b
11 = b

so ur perpendicular equation is : y = -x + 11 <==
8 0
3 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.

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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 < ∞.

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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.

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2 years ago
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3,503,832.97 hope this helps! :)
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I hope this helps :D

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