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Nezavi [6.7K]
3 years ago
12

Which congruence theorem can be used to prove bda = bdc Hlssa aas sss

Mathematics
2 answers:
Crank3 years ago
7 0

Answer: the answer is sss

Step-by-step explanation:

ozzi3 years ago
4 0
Given that ΔBDA is similar to ΔBDC and:
AD≡DC
AB≡BC
BD≡BD (shared side)
then the best postulate to use is the side-side-side (SSS) postulate.
Answer: SSS
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Which of the following could be the function
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What is the length of side x?
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11 because as you see 3 and 6 if you divide it the scale factor is 1/2 so you just do 5.5 times 2 and get 11

Step-by-step explanation:

6 0
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Read 2 more answers
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
Sin (5x-25) = COS (3x)<br> X=?
Advocard [28]

Answer:

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Step-by-step explanation:

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8 0
1 year ago
(2x - 3y + 1) + (4x + y + 5)=​
disa [49]
6x-2y+5. Here you just do addition/subtraction with the coefficients on the variables.
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