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Sergeeva-Olga [200]
2 years ago
8

Two exponential functions are shown in the table.

Mathematics
1 answer:
sammy [17]2 years ago
5 0

Based on the table, a conclusion which can be drawn about f(x) and g(x) is that: B. the functions f(x) and g(x) are reflections over the y-axis.

<h3>How to compare the functions f(x) and g(x)?</h3>

In Mathematics, two functions are considered to be reflections over the y-axis under the following condition:

If, f(-x) = g(x).

Evaluating the given functions, we have:

f(x) = 2ˣ

f(-x) = 2⁻ˣ = ½ˣ = g(x).

Similarly, two functions are considered to be reflections over the x-axis under the following condition:

If, -f(x) = g(x).

Evaluating the given functions, we have:

f(x) = 2ˣ

-f(x) = -2ˣ ≠ g(x).

Therefore, we can logically conclude that the two functions f(x) and g(x) are considered to be reflections over the y-axis but not the x-axis.

Read more on reflections here: brainly.com/question/2702511

#SPJ1

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Mateo reproducira una foto que mide 5 cm por lado a una escala de 3/4
Aneli [31]

the question in English

Mateo will reproduce a photo measuring 5 cm per side on a scale of 3/4

How much will the square side of the copy measure?

Will the size of the copy be larger or smaller than the original?

we know that

the scale factor is equal to \frac{3}{4}

Let

x----------> the original length side

y--------> the copy length side

y=\frac{3}{4}*x

we have that

x=5\ cm

substitute the values

y=\frac{3}{4}*5=3.75\ cm

therefore

<u>the answer Part a) is </u>

the length side of the copy measure is 3.75\ cm

Part b)

we know that

the size of the original photo is------> 5*5=25\ cm^{2}

the size of the copy is------> 3.75*3.75=14.06\ cm^{2}

so

14.06\ cm^{2} < 25\ cm^{2}

therefore

<u>the answer part b) is</u>

The size of the copy be smaller than the original



3 0
3 years ago
-3( 2u + 5 ) +3.5u = -1u solve the equation
lora16 [44]

Answer:

that conclude idk

Step-by-step explanation:

-6U+-15+3.5U=-1U

-2.5U+15

4 0
3 years ago
What set of Reflections and rotations could carry ABCD onto itself?​
Nitella [24]

Reflect over y-axis,reflect over the X axis ,rotate 180°

Option D is the correct option.

<em>Expla</em><em>nation</em><em>:</em>

<em>Let's </em><em>take</em><em> </em><em>point</em><em> </em><em>A</em><em> </em><em>which </em><em>is</em><em> </em><em>(</em><em>4</em><em>,</em><em>-</em><em>1</em><em>)</em>

<em>Reflect</em><em>ion</em><em> </em><em>over</em><em> </em><em>y-</em><em> </em><em>axis </em><em>will</em><em> </em><em>make</em><em> </em><em>this</em><em> </em><em>point</em><em> </em><em>(</em><em>4</em><em>,</em><em>1</em><em>)</em>

<em>Then</em><em>,</em><em> </em><em>reflect</em><em>ion</em><em> </em><em>over</em><em> </em><em>X </em><em>axis</em><em> </em><em>will</em><em> </em><em>make</em><em> </em><em>this</em><em> </em><em>point </em><em>(</em><em>4</em><em>,</em><em>-</em><em>1</em><em>)</em>

<em>After</em><em> </em><em>rota</em><em>tion</em><em> </em><em>of</em><em> </em><em>1</em><em>8</em><em>0</em><em> </em><em>degree</em><em> </em><em>we</em><em> </em><em>will</em><em> </em><em>get</em><em> </em><em>(</em><em>-</em><em>4</em><em>,</em><em>1</em><em>)</em><em> </em><em>.</em>

<em>Please</em><em> </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>Hope</em><em> </em><em>it </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>

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3 years ago
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An observer in a hot air balloon sights a building that is 50 m from the balloon's launch point. The balloon has risen 165 m. Wh
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Notice the picture,
recall your SOH, CAH, TOA
\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad \qquad &#10;% cosine&#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;&#10;\\ \quad \\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}

you have,
opposite side, 165
adjacent side, 50
and the angle

that means, we'll need Mrs. tangent
thus 
\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50}&#10;\\ \quad \\&#10; tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right]&#10;\\ \quad \\&#10;\theta=tan^{-1}\left[ \cfrac{165}{50}\right]&#10;\\ \uparrow  \\&#10; \textit{angle of elevation}\iff\textit{angle of depression}

4 0
3 years ago
I need help on this. I don’t get how to do it
-BARSIC- [3]
<h3>Answer:</h3>

I haven't learn that

7 0
3 years ago
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