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madreJ [45]
3 years ago
15

Simplify √(x^2-10x+25) if -5≤x<5

Mathematics
1 answer:
romanna [79]3 years ago
8 0

\qquad\qquad\huge\underline{{\sf Answer}}

Let's simplify ~

\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 10x + 25 }

\qquad \sf  \dashrightarrow \: \sqrt{  {x}^{2} - 5x - 5x   + 25  }

\qquad \sf  \dashrightarrow \: \sqrt{  {x}^{} (x- 5) - 5(x    - 5)  }

\qquad \sf  \dashrightarrow \: \sqrt{   (x- 5) (x    - 5)  }

\qquad \sf  \dashrightarrow \: \sqrt{   (x- 5)  {}^{2}  }

\qquad \sf  \dashrightarrow \: (x- 5)

value x lies between :

\qquad \sf  \dashrightarrow \: - 5 \leqslant x \leqslant 5

if the value of x is taken -5

\qquad \sf  \dashrightarrow \: - 5 - 5

\qquad \sf  \dashrightarrow \: - 10

if value of x is taken as 5

\qquad \sf  \dashrightarrow \:5 - 5

\qquad \sf  \dashrightarrow \:0

So, the possible values of the required expression lies between ~

\qquad \sf  \dashrightarrow \: - 10 \leqslant  \sqrt{ {x}^{2} - 10x + 25 }  < 0

I hope you understood the whole procedure. let me know if you have any doubts in given steps ~

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Find the unit vectors that are parallel to the tangent line to the parabola y = x2 at the point (2, 4).
skelet666 [1.2K]

Answer:

The unit vectors are:

\vec u_1 = \left\\\vec u_2 = \left

Step-by-step explanation:

Unit vectors that are a parallel to the tangent line have the same slope than the tangent line, thus we can find the slope of the tangent line, find directional vectors and then their corresponding unit vectors.

Finding the slope of the tangent line.

We can find the first derivative of the curve, which represents the slope of the tangent line of the curve at that point.

y = x^2 \\ y'=2x

At x = 2 we have

y'= 2(2)\\y'= 4\\

Thus we have slope m = 4, then the parallel unit vector has slope m = 4/1

Finding unit vectors.

From the slope m = 4/1, we can write it as a direction vector with x =1 and y = 4. Notice also that x = -1 and y = -4 would have given as as well slope m = 4 too.

\vec u =

Then in order to find the unit vector on that direction we can use the formula

\hat u =\cfrac{\vec u}{|\vec u|}

So finding the magnitude we get

|\vec u |= \sqrt{(1)^2+(4)^2}\\|\vec u| = \sqrt{17}

Then one unit vector that is parallel to the tangent line is

\vec u_1 = \left

And the second unit vector is

\vec u_2 = \left

The negative sign on both x and y components just tell us that we are aiming on the opposite direction as the first unit vector, yet both have the same value of the slope.

6 0
3 years ago
Write the equation for a parabola with a focus at (1,-4)(1,−4)left parenthesis, 1, comma, minus, 4, right parenthesis and a dire
lutik1710 [3]

Answer:

x = 3/2 - ((y + 4)²)/2

Step-by-step explanation:

The coordinates of the focus of the parabola = (1, -4)

The directrix of the parabola is the line, x = 2 (parallel to the y-axis)

The general form of a parabola having a directrix parallel to the y-axis is presented as follows;

(y - k)² = 4·p·(x - h)

The coordinates of the focus = (h + p, k)

The line representing the directrix is x = h - p

Therefore, by comparison, we have;

(1, -4) = (h + p, k)

k = -4

h + p = 1

∴ p = 1 - h

From the directrix, we have;

2 = h - p

∴ 2 = h - (1 - h) = 2·h - 1

2·h = 2 + 1

h = 3/2

p = 1 - h

∴ p = 1 - 3/2 = -1/2

p = -1/2

By substitution of the values of <em>k</em>, <em>p</em>, and <em>h</em>, in the general equation of the parabola, (y - k)² = 4·p·(x - h), we get;

(y - k)² = 4·p·(x - h) = (y - (-4))² = 4·(-1/2)·(x - (3/2)) = (y + 4)² = -2·(x - 3/2)

The equation of the parabola in vertex form is therefore;

x = (-1/2)·(y + 4)² + 3/2 = 3/2 - (y + 4)²/2

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3 years ago
Solve in radians help plz
PIT_PIT [208]
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cos x -  sqrt3 sqrt ( 1 - cos x) /sqrt2 = 1

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sqrt(3/2)(sqrt(1 - cos x) =  cos x - 1   Squaring both sides:-
1.5 ( 1 - cos x) = cos^2 x - 2 cos x + 1

cos^2 x - 0.5 cos x - 0.5 = 0 

cos x = 1 , -0.5

giving x = 0 , 2pi, 2pi/3,  4pi/3  ( for  0  =< x <= 2pi)

because of thw square roots some of these solutions may be extraneous so we should plug these into the original equations to see if they fit.

The last 2 results dont fit so the answer is  x = 0 , 2pi Answer
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The results of a survey are shown in the pie chart below and the survey 28 students say their favorite dessert is ice cream how
Rainbow [258]

Answer:

12 students.

Step-by-step explanation:

The complete question would be:

If ice cream is the last dessert, then we can solve for how many percent of the students like ice cream. If we add the 4 desserts percent, it should be equal to 100%

Pie : 15%

Candy : 23%

Cake : 27%

Pie% + Candy% + cake% + ice cream% = 100%

15% + 23% + 27% + Ice cream% = 100%

65% + Ice cream% = 100%

Ice cream% = 100% - 65%

Ice cream % = 35%

Percent is a way of saying how many out of a 100. Before we can tell how many students like pie, we need to figure out how many students were surveyed.

It says here that 35% of them like ice cream and among the students, 28 students like ice cream. This means that 28 is 35% of a number, which will be the number of students surveyed.

So as a fraction:

35\% = \dfrac{35}{100}

We need to find a ratio that is proportional to that, with 28 as the part of the whole.

\dfrac{35}{100} = \dfrac{28}{x}\\\\Cross-multiply\\\\35x = 28(100)\\\\35x = 2800\\\\Divide\;both\;sides\;by\;35\\\\\dfrac{35x}{35} = \dfrac{2800}{35}\\\\x=80

So that is the number of students survey. Now to look for how may people like pie, we find what is 15% of 80. So we find the part of the whole this time:

\dfrac{15}{100}=\dfrac{x}{80}\\\\15(80)=100x\\\\1200 = 100x\\\\\dfrac{1200}{100}=\dfrac{100x}{100}\\\\12 = x

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