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3 years ago

The mirror image of the point (2, 5) in the x – axis is?

Mathematics

2 answers:

zubka84 [21]3 years ago

5 0

Answer:

(2,-5).

Step-by-step explanation:

You see, when you reflect over something, one variable has to change. Our line of reflection is the x-axis, or y=0.

Here is a rule to help with these kinds of problems.

If you reflect over the x-axis, the x value is still the same, but the y value changes to be the inverse or opposite value. The opposite of positive is negative, so it will be -5.

For reflected over the y-axis, the y value stays the same, but the x value changes to be the inverse of the original y value.

Alekssandra [29.7K]3 years ago

4 0

Answer:

(2,-5)

Step-by-step explanation:

Reflecting (2,5) across x-axis will invert it's y-coordinate and will make it negative.

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Meg initially has 3 hours of pop music and 2 hours of classical music in her collection. Every month onwards, the hours of pop m

zlopas [31]

Answer: Option 'd' is correct.

Step-by-step explanation:

Since we have given that

Number of hours of pop music = 3

Number of hours of classical music = 2

According to question, Every month onwards, the hours of pop music in her collection is 5% more than what she had the previous month. Her classical music does not change.

Rate of increment = 5% = 0.05

Let the number of months be 'x'.

So, our required function becomes,

$f(x)=3(1+0.05)x+2\\\\f(x)=3(1.05)x+2$

Hence, Option 'd' is correct.

3 0

2 years ago

What is the distance between the points (2,10) and (-6, 4) on the coordinate

stellarik [79]

Answer:

The distance between the given points (2,10) and (-6, 4) on the coordinate plane is 10units

Therefore distance s=10 units

Step-by-step explanation:

Given points are (2,10) and (-6, 4) on the coordinate plane

To distance between the given points :

The distance formula is $s=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ units

Let $(x_1,y_1)$ , $(x_2,y_2)$ be the given points (2,10) and (-6, 4) respectively

Now substituting the values in the distance formula we get

$s=\sqrt{(-6-2)^2+(4-10)^2}$

$=\sqrt{(-8)^2+(-6)^2}$

$=\sqrt{8^2+6^2}$

$=\sqrt{64+36}$

$=\sqrt{100}$

$=10$

Therefore s=10 units

The distance between the given points (2,10) and (-6, 4) on the coordinate plane is 10units

5 0

3 years ago

Which statement below is correct

Lynna [10]

The correct answer is A

4 0

3 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$

8 0

2 years ago

Janet deposited $9,600 into an account that pays 4.4 interest, compounded daily. At the end of nine months, how much interest ha

neonofarm [45]

bearing in mind that 9 months is not even a year, but since there are 12 months in a year, then 9 months is really 9/12 years.

$\bf ~~~~~~ \stackrel{\textit{daily}}{\textit{Continuously}} \textit{Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$9600\\ r=rate\to 4.4\%\to \frac{4.4}{100}\dotfill &0.044\\ t=years\to \frac{9}{12}\dotfill &\frac{3}{4} \end{cases} \\\\\\ A=9600e^{0.044\cdot \frac{3}{4}}\implies A=9600e^{0.033}\implies A\approx 9922.09 \\\\\\$

$\bf \stackrel{\textit{interest earned}}{9922.09-9600\implies 322.09}$

5 0

3 years ago