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Vera_Pavlovna [14]
3 years ago
12

This one please. Thank you!!

Mathematics
2 answers:
lys-0071 [83]3 years ago
5 0
I’m believe it’s the 3rd one but i’m not sure
Ilya [14]3 years ago
5 0

Answer:

The answer highlighted in yellow in your picture looks like the correct answer to me!

Step-by-step explanation:

hope this is helpful!

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Jina is mailing packages. Each small package costs her to send. Each large package costs her . How much will it cost her to send
Novosadov [1.4K]
You need to tell me how many large and small packages she’s sending and how much each costs
8 0
3 years ago
Answer this plz, i couldn't figure this out.
AfilCa [17]

Answer:

Step-by-step explanation:

Mean - “Average” Mean teachers average grades.

Add all numbers up

Divide by how many numbers you added.

Median - “Middle” The median of the road is the middle lane.

Put the numbers in order. (Least to Greatest)

Find the middle number.

IF there are two numbers in the middle add them up and divide by two.

Mode - “Most” The number that appears the most.

Put the numbers in order. (Least to Greatest)

Find the most repeated number.

IF none of the numbers repeat there is no mode.

Range - “Spread” Subtract the highest number and the lowest number.

Put the numbers in order. (Least to Greatest)

Subtract the highest number and the lowest number.

Minimum - the first point, the smallest number.

Lower Quartile - the second point, find the median of the minimum and median.

Median - the third point, find the middle number.

Upper Quartile - the fourth point, find the median of the median and maximum.

Maximum - the fifth point, the largest number.

Spread - (Range) subtract the max and the min.

Interquartile Range (IQR) - range of the box, subtract the upper and the lower.

Composite

A composite number is a positive integer that can be formed by multiplying two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.

Median

If there is an odd number of data points, then you will have just one middle number. If there is an even number of data points, then you need to pick the two middle numbers, add them together, and divide by two. That number will be your median. Mode - The mode is the number that appears the most

5 0
3 years ago
Read 2 more answers
Show all work to identify the asymptotes and zero of the function f(x) = 6x / x^2 - 36
eduard

Answer:

Zero of the function f(x) is at x = 0

Vertical Asymptotes at x = ±6

Horizontal Asymptotes at y = 0

Step-by-step explanation:

<h3>Vertical Asymptotes </h3>

For a given function f(x):

Vertical Asymptotes are obtained at those values of x, where the function f(x) tends to infinity, I.e.,

<em>When</em><em> </em><em>x</em><em> </em><em>approaches</em><em> </em><em>some</em><em> </em><em>constant</em><em> </em><em>value</em><em> </em><em>b</em><em>u</em><em>t</em><em> </em><em>th</em><em>e</em><em> </em><em>curve</em><em> </em><em>moves</em><em> </em><em>towards</em><em> </em><em>infinity</em><em>.</em><em> </em>

  • If f(x) is a fraction, it'll tend to infinity when it's denominator becomes zero.

Vertical Asymptotes of the given function can be obtained by walking thru the following steps:

<u>Step I</u>

(Factorise the numerator and denominator)

\mathsf{ f(x) = \frac{6x}{ {x}^{2} - 36 } }

<em>x</em><em>²</em><em> </em><em>-</em><em> </em><em>36</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>facto</em><em>rised</em><em> </em><em>into</em><em> </em><em>(</em><em>x</em><em> </em><em>+</em><em> </em><em>6</em><em>)</em><em>(</em><em>x</em><em> </em><em>-</em><em> </em><em>6</em><em>)</em>

<em>and</em><em>,</em><em> </em><em>ofcourse</em><em>,</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>write</em><em> </em><em>6</em><em>x</em><em> </em><em>as</em><em> </em><em>6</em><em>(</em><em>x</em><em> </em><em>-</em><em> </em><em>0</em><em>)</em><em> </em>

\mathsf{ f(x) = \frac{6(x - 0)}{ (x + 6)(x - 6) } }

<u>Step</u><u> </u><u>II</u>

(Reduce the fraction to its simplest form by canceling out the common factors)

<em>There aren't any common factors in the numerator and denominator in this case.</em>

<u>Step</u><u> </u><u>III</u>

(Look for the values of x which cause the denominator to be zero)

<em>If</em><em> </em><em>we</em><em> </em><em>put</em><em> </em>x = 6

<em>denominator</em><em> </em><em>becomes</em><em> </em><em>0</em>

Also,

<em>If</em><em> </em><em>we</em><em> </em><em>substitute</em><em> </em><em>x</em><em> </em><em>with</em><em> </em> -6

<em>denominator</em><em> </em><em>becomes</em><em> </em><em>0</em><em>.</em><em> </em>

The two values of x indicate the two Vertical Asymptotes of the function f(x).

Therefore,

<u>Vertical</u><u> </u><u>Asymptotes</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>function</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u><u>are</u><u>:</u>

\boxed{ \mathsf{x =  \pm6}}

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3 /><h3>Horizontal Asymptotes:</h3>

Horizontal Asymptotes are obtained When x tends to infinity and y approaches some constant value.

I'll be using the concept of limits for this.

\mathsf{y = \frac{6x}{ {x}^{2} - 36 }  }

<em>dividing</em><em> </em><em>and</em><em> </em><em>multiplying</em><em> </em><em>by</em><em> </em><em>x</em><em>²</em><em> </em><em>(</em><em>Yep</em><em>!</em><em> </em><em>so</em><em> </em><em>if</em><em> </em><em>x</em><em> </em><em>becomes</em><em> </em><em>infinity</em><em> </em><em>1</em><em>/</em><em> </em><em>x</em><em> </em><em>and</em><em> </em><em>1</em><em>/</em><em> </em><em>x</em><em>²</em><em> </em><em>all</em><em> </em><em>such</em><em> </em><em>terms</em><em> </em><em>become</em><em> </em><em>0</em><em>,</em><em> </em><em>'</em><em>cause</em><em> </em><em>1</em><em>/</em><em> </em><em>∞</em><em> </em><em>is</em><em> </em><em>0</em><em>)</em><em> </em>

\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6x}{ {x}^{2} } }{  \frac{ {x}^{2} - 36 }{ {x}^{2} }  } ) }

\implies \mathsf{y = lim_{x \rightarrow \infty }( \frac{ \frac{6}{ x } }{  1-  \frac{36 }{ {x}^{2} }  } ) }

Substitute x with ∞, you get zero/ 1

\implies  \boxed{\mathsf{y = 0}}

So, the horizontal Asymptote of the function is y = 0, that is the x axis

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3>Zeroes of a function:</h3>

The values of x that reduces f(x) to zero are called the zeroes of f(x).

Here, only x = 0 acts as the zero of the function.

[NOTE:

  • For finding <u>Vertical Asymptotes</u><u>,</u>Equate the denominator to 0. And
  • For finding <u>Zeroes</u><u>,</u> Equate the numerator to 0]

__________________

[That's what it's graph looks like. ]

3 0
3 years ago
I have no idea how to do this, someone help me
julsineya [31]
Try this:
1) for square a=5*2=10.
for circle r=5.
2) for square S'=10²=100; for circle S°=πr²≈78.54.
3) for half of circle S₁=0.5*78.54=39.27
4) for shaded region S=S'-S₁=100-39.27≈60.73
Answer: G (60.73)
6 0
3 years ago
Jim designed a rectangular garden with a width of 15 feet. The area of the garden is 300 square feet. What is the length of the
Elodia [21]

Answer:

Length of the rectangle = 20ft

Step-by-step explanation:

A rectangle is a quadrilateral that is a four sided polygon with two opposite sides parallel and equal.

Area of a rectangle = length x width

From question

Area = 300ft² , width = 15 ft,  length = ?

Input values.

Area = L x w

300 = L × 15

15L = 300

Divide both sides by 15

15L/15 = 300ft²/15ft

L = 20 ft

Length of the rectangle = 20ft

I hope this was helpful, please rate as brainliest

4 0
3 years ago
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