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ankoles [38]
3 years ago
10

Help! locus points!

Mathematics
1 answer:
Liula [17]3 years ago
5 0

Answer:

  1. Circle X with radius 2 cm.
  2. Either of two lines parallel to AB.

Step-by-step explanation:

1. The definition of a circle is all the points in a plane that are at some radius r from a given point (the circle center). That is what you have, with a radius of 2 cm.

__

2. Parallel lines are the same distance apart everywhere. Each line will have two parallel lines at some given distance from it, one on each side. Here, the separation distance from AB is 1 cm, so your locus of points is the two lines parallel AB that are 1 cm from it on either side.

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PLEASE ANSWER THIS? WHAT IS EG
crimeas [40]
It's the points of the triangle E----------G
They are the twin points
5 0
2 years ago
In the diagram below, O is circumscribed about quadrilateral DEFG. What is the value of x?
Fofino [41]

Answer:

A. 70

Step-by-step explanation:

The given quadrilateral DEFG is a cyclic quadrilateral.

\angle F+70\degree=180\degree...opposite angles of a cyclic quadrilateral are supplementary.

\implies \angle F=180\degree-70\degree

\implies \angle F=110\degree

The sum of angles in a quadrilateral is 360 degrees.

x-10+70+110+120=360

x+290=360

x=360-290

x=70\degree

7 0
3 years ago
What is the equation of a parabola with a directrix of y=2 and a focus point of 0,-2
KiRa [710]
Hope this helped. :)

Any point, <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> on the parabola satisfies the definition of parabola, so there are two distances to calculate:

<span>Distance between the point on the parabola to the focusDistance between the point on the parabola to the directrix</span>

To find the equation of the parabola, equate these two expressions and solve for <span><span>y0</span><span>y0</span></span> .

Find the equation of the parabola in the example above.

Distance between the point <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> and <span><span>(<span>a,b</span>)</span><span>(<span>a,b</span>)</span></span> :

<span><span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span><span>‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾</span>√</span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span></span>

Distance between point <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> and the line <span><span>y=c</span><span>y=c</span></span> :

<span><span><span>∣∣</span><span><span>y0</span>−c</span><span>∣∣</span></span><span>| <span><span>y0</span>−c</span> |</span></span>

(Here, the distance between the point and horizontal line is difference of their <span>yy</span> -coordinates.)

Equate the two expressions.

<span><span><span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span><span>‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾</span>√</span>=<span><span>∣∣</span><span><span>y0</span>−c</span><span>∣∣</span></span></span><span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span></span>=<span>| <span><span>y0</span>−c</span> |</span></span></span>

Square both sides.

<span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span>=<span><span>(<span><span>y0</span>−c</span>)</span>2</span></span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span><span>(<span><span>y0</span>−b</span>)</span>2</span>=<span><span>(<span><span>y0</span>−c</span>)</span>2</span></span></span>

Expand the expression in <span><span>y0</span><span>y0</span></span> on both sides and simplify.

<span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span>b2</span>−<span>c2</span>=2<span>(<span>b−c</span>)</span><span>y0</span></span><span><span><span>(<span><span>x0</span>−a</span>)</span>2</span>+<span>b2</span>−<span>c2</span>=2<span>(<span>b−c</span>)</span><span>y0</span></span></span>

This equation in <span><span>(<span><span>x0</span>,<span>y0</span></span>)</span><span>(<span><span>x0</span>,<span>y0</span></span>)</span></span> is true for all other values on the parabola and hence we can rewrite with <span><span>(<span>x,y</span>)</span><span>(<span>x,y</span>)</span></span> .

Therefore, the equation of the parabola with focus <span><span>(<span>a,b</span>)</span><span>(<span>a,b</span>)</span></span> and directrix <span><span>y=c</span><span>y=c</span></span> is

<span><span><span><span>(<span>x−a</span>)</span>2</span>+<span>b2</span>−<span>c2</span>=2<span>(<span>b−c</span>)</span>y</span></span>

3 0
3 years ago
A family wants to save for college tuition for their daughter. What continuous yearly interest rate r% is needed in their saving
sergey [27]

Answer:

The continuous yearly interest is  22.5% per year.

Step-by-step explanation:

Continuous yearly interest:

Continuous yearly interest is defined as the sum of the interest comes from principle and the interest comes from interest.

The formula for continuous interest yearly is

A=Pe^{rt}

where A = The final amount =$110,000

P= principle =$4,700

r= rate of interest

t= time (in year)= 14 years

\therefore 110,000= 4,700e^{r\times 14}

\Rightarrow e^{14r}= \frac{110,000}{4,700}

Taking ln both sides

\Rightarrow ln e^{14r}= ln(\frac{110,000}{4,700})

\Rightarrow {14r}= ln(\frac{1100}{47})

\Rightarrow r=\frac{ln( \frac{1100}{47})}{14}

\Rightarrow r = 0.225  (approx)

The continuous yearly interest is 0.225 = 22.5% per year.

4 0
3 years ago
Hello, am I pretty?
vichka [17]

Answer:

wdym

Step-by-step explanation:

8 0
2 years ago
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