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babunello [35]
3 years ago
12

What is the range of the function f(x) = –2(6x) + 3?

Mathematics
1 answer:
padilas [110]3 years ago
5 0

Answer: y=-12x+3                         hope it helps

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Pls help me with A, B, C
Mariulka [41]

Answer:

a. is 3

Step-by-step explanation:

(1/3)* (6+3) = 3

6 0
3 years ago
Find the difference between 7 1/2 and 2 3/4
NNADVOKAT [17]

Answer:

4\frac{3}{4} or 4.75

Step-by-step explanation:

first we're going to convert the number into improper fractions.

7\frac{1}{2}= \frac{15}{2} and 2\frac{3}{4}= \frac{11}{4}

For subtracting it is suitable to adjust both fractions to an equal denominator, so we're going to multiply \frac{15}{2}x2, it'll equal the same number.

So now we have \frac{30}{4}-\frac{11}{4}, 30-11= 19.

\frac{19}4} simplified is, 4\frac{3}{4}.

As you can see I also added 4.75 as an answer, that's if the answer must be a decimal.

I hope this helps you!

3 0
3 years ago
The complex solution to a quadratic equation is x equals start fraction three plus or minus square root of negative 36 end squar
Naya [18.7K]
The square root of a -36 simplifies to "6i". So you have 3+/- [6i/6].  The 6's cancel each other out leaving 3+/-i.  So the "a" is 3 and the "b" is 1
4 0
3 years ago
I need the answer for 90 please its due tommorow!
babunello [35]
Jacqueline is incorrect. 
To find how much trash you can clean in 4 days if you already know how much trash you can clean in one day, multiply 1 2/3 (how much you can clean in one day) by 4. 
Another way to write 1 and 2/3 miles is 5/3 (converting a mixed number to a single fraction). Then, 5/3 x 4 = 20/3. (To multiply fractions, just multiply across, leaving 5 x 4 in the numerator and 4 x 1 in the denominator.)
Jacqueline said you would be able to clean 19/3 miles but you can actually clean 20/3.
4 0
3 years ago
Read 2 more answers
Both the La Plata river dolphin (Pontoporia blainvillei) and
Citrus2011 [14]

Answer:

<em>1. A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin; 2. The radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball; 3. The volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

Step-by-step explanation:

If we expressed a number as:

\\ N = a * 10^{b} (1)

Where

\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10} (2)

or

\\ 1 \leq a < 10 (3)

Then, <em>b</em> represents the <em>order of magnitude </em>of such a number (<em>Order of magnitude (2020), </em>in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The<em> La Plata river dolphin</em> weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the <em>scientific notation</em>):

\\ 30kg \leq Dolphin_{weight} \leq 50kg

\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg

\\ 35000kg \leq Whale_{weight} \leq 40000kg

\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)

\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)

Then

\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}

Thus

<em>A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin.</em>

<h3>Second case</h3>

Following the same reasoning, we can conclude that <em>the radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball:</em>

\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}

\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}

<h3>Third case</h3>

For this case, we need to calculate <em>the volume of a sphere</em> for both radii (1cm and 10cm).

The volume of a sphere is

\\ V_{sphere} = \frac{4}{3}*\pi*R^{3}

Then, the volume of the <em>ball of radius 1cm</em> is:

\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}

\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}

And, the volume of the <em>ball of radius 10cm</em> is:

\\ V_{radius=10} = \frac{4}{3}*\pi*(10cm)^{3}

\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}

Thus

\\ \frac{10^{3}}{10^{0}} = 10^{3}

As a result, <em>the volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

4 0
3 years ago
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