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k0ka [10]
3 years ago
13

Determine whether each function represents exponential growth or decay. Select the correct option for each function. PLEASE HELP

ASAP

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer: f(x)= (3/4)^x/2 —> Decay

f(x)=(4/3)^x —> Growth

f(x)=(5/6)^3x —> Decay

f(x)=(8/3)^x/3 —> Growth

f(x)=(3/2)^2x —> Growth

Step-by-step explanation: When you graph them one by one, for example, the first equation (f(x)= (3/4)^x/2), the graph would have an asymptote on the right side, meaning that it’s decaying.

Straight up explanation: if the asymptote on the graph is on the left then it is growing, but if the asymptote is on the right then it’s decaying.

Asymptote is the “line” that almost touches the x axis.

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Answer:

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7 0
3 years ago
Rewrite 5\11 and 3\7 so that they have common denominator
Veseljchak [2.6K]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
Si un kilogramo de arroz cuesta 24.50,cuanto se pagara por 0.75 kg y 4 kg?
klemol [59]

Answer:

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Step-by-step explanation:

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3 years ago
What else would need to be congruent to show that AABC=A DEF by the
Pepsi [2]

Answer:

D. AC ≅ DF

Step-by-step explanation:

According to the AAS Theorem, two triangles are considered congruent to each other when two angles and a mon-included side of one triangle are congruent to two corresponding angles and a corresponding non-included side of the other.

Thus, in the diagram given:

<A and <B in ∆ABC are congruent to corresponding angles <D and <E in ∆DEF.

The only condition left to be met before we can conclude that both triangles are congruent by the AAS Theorem is for a mon-included side AC to be congruent to corresponding non-included side DF.

So, AC ≅ DF is what is needed to make both triangles congruent.

7 0
3 years ago
I am trying to find the equation of the following lines, i’m just having trouble putting them in slope-intercept form.
Margaret [11]
\bf \begin{array}{ccccccccc}&#10;&&x_1&&y_1\\&#10;&&(~ -1 &,& -6~)&#10;\end{array}&#10;\\\\\\&#10;slope =  m\implies  \cfrac{2}{5}&#10;\\\\\\&#10;\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-6)=\cfrac{2}{5}[x-(-1)]&#10;\\\\\\&#10;y+6=\cfrac{2}{5}(x+1)\implies y+6=\cfrac{2}{5}x+\cfrac{2}{5}&#10;\\\\\\&#10;y=\cfrac{2}{5}x+\cfrac{2}{5}-6\implies y=\cfrac{2}{5}x-\cfrac{28}{5}



\bf \begin{array}{ccccccccc}&#10;&&x_1&&y_1&&x_2&&y_2\\&#10;&&(~ -2 &,& 5~) &#10;&&(~ 5 &,& 8~)&#10;\end{array}&#10;\\\\\\&#10;slope =  m\implies &#10;\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{8-5}{5-(-2)}\implies \cfrac{8-5}{5+2}\implies \cfrac{3}{7}&#10;\\\\\\&#10;\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-5=\cfrac{3}{7}[x-(-2)]&#10;\\\\\\&#10;y-5=\cfrac{3}{7}(x+2)\implies y-5=\cfrac{3}{7}x+\cfrac{6}{7}&#10;\\\\\\&#10;y=\cfrac{3}{7}x+\cfrac{6}{7}+5\implies y=\cfrac{3}{7}x+\cfrac{41}{7}
7 0
3 years ago
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