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Ierofanga [76]
3 years ago
14

In a reflection along the x-axis a given coordinate (2, -1) transforms itself into which of the following?

Mathematics
2 answers:
Mariulka [41]3 years ago
7 0
If that coordinate is reflected along the x-axis, the image will be (2,1).
katovenus [111]3 years ago
5 0

Answer with explanation:

The Meaning of reflection of a point through any of the axis states that,the perpendicular distance of a point from either of the axes chosen as line of reflection from Image and Preimage is constant.

⇒Preimage = (2, -1)

Marking the point in x y Plane and then Plotting it's reflection in x y plane

⇒ Image = (2, 1)

In the first Quadrant.

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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

5 0
2 years ago
If human height were quantized in one-foot increments, what would happen to the height of a child as she grows up?
agasfer [191]
I believe what the problem means is that the height of the child, if he/she grows in height, the mesurements are equal to 1 ft, 2 ft, 3 ft, 4 ft, etc only. and no decimals or fractions in between. it is also said that a human stops growing at the age of 20 naturally, hence with this increment, a child can grow incredibly very tall.
7 0
3 years ago
Write 425% as a fraction or mixed number in simplest form.
Temka [501]

Answer:

17/40

here! hope this helps!

8 0
2 years ago
Please help! provide thorough explanation and use elimination method!!
ICE Princess25 [194]

Multiply both sides of the second equation by 100 to get rid of the decimals:

0.05<em>n</em> + 0.10<em>d</em> = 1.50

==>  5<em>n</em> + 10<em>d</em> = 150

Multiply both sides of the first equation by -5:

<em>n</em> + <em>d</em> = 21

==>  -5<em>n</em> - 5<em>d</em> = -105

Add the two equations together:

(5<em>n</em> + 10<em>d</em>) + (-5<em>n</em> - 5<em>d</em>) = 150 + (-105)

Notice that the terms containing <em>n</em> get eliminated and we can solve for <em>d</em> :

(5<em>n</em> - 5<em>n</em>) + (10<em>d</em> - 5<em>d</em>) = 150 - 105

5<em>d</em> = 45

<em>d</em> = 45/5 = 9

Plug this into either original equation to solve for <em>n</em>. Doing this with the first equation is easiest:

<em>n</em> + 9 = 21

<em>n</em> = 21 - 9 = 12

So Donna used 12 nickels and 9 dimes.

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3 years ago
How do you go from slope-intercept form to standard form
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Thats How you go from slope to to standard form

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