All categories

1 year ago

which of the following are among the five basic postulates of euclidean geometry? check all that apply

1 answer:

7 0

The five essential hypothesizes of Geometry, additionally alluded to as Euclid's proposes are the accompanying:
1.) A straight line section can be drawn joining any two focuses.
2.) Any straight line portion can be expanded uncertainly in a straight line.
3.) Given any straight line fragment, a circle can be drawn having the portion as a span and one endpoint as the inside.
4.) All correct points are harmonious.
5.) If two lines are drawn which meet a third such that the total of the internal points on one side is under two right edges (or 180 degrees), then the two lines unavoidably should converge each other on that side if reached out sufficiently far.

You might be interested in

(i)Express 2x² – 4x + 1 in the form a(x+ b)² + c and hence state the coordinates of the minimum point, A, on the curve y= 2x² 4x

earnstyle [38]

Answer:

(i). $y = 2\, x^2 - 4\, x + 1 = 2\, (x - 1)^2 - 1$ . Point $A$ is at $(1, \, -1)$ .

(ii). Point $Q$ is at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ .

(iii). $\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$ (slope-intercept form) or equivalently $x + 5\, y - 17 = 0$ (standard form.)

Step-by-step explanation:

<h3>Coordinates of the Extrema</h3>

Note, that when $a(x + b)^2 + c$ is expanded, the expression would become $a\, x^2 + 2\, a\, b\, x + a\, b^2 + c$ .

Compare this expression to the original $2\, x^2 - 4\, x + 1$ . In particular, try to match the coefficients of the $x^2$ terms and the $x$ terms, as well as the constant terms.

For the $x^2$ coefficients: $a = 2$ .
For the $x$ coefficients: $2\, a\, b = - 4$ . Since $a = 2$ , solving for $b$ gives $b = -1$ .
For the constant terms: $a \, b^2 + c = 1$ . Since $a = 2$ and $b = -1$ , solving for $c$ gives $c =-1$ .

Hence, the original expression for the parabola is equivalent to $y = 2\, (x - 1)^2 - 1$ .

For a parabola in the vertex form $y = a\, (x + b)^2 + c$ , the vertex (which, depending on $a$ , can either be a minimum or a maximum,) would be $(-b,\, c)$ . For this parabola, that point would be $(1,\, -1)$ .

<h3>Coordinates of the Two Intersections</h3>

Assume $(m,\, n)$ is an intersection of the graphs of the two functions $y = 2\, x^2- 4\, x + 1$ and $x -y + 4 = 0$ . Setting $x$ to $m$ , and $y$ to $n$ should make sure that both equations still hold. That is:

$\displaystyle \left\lbrace \begin{aligned}& n = 2\, m^2 - 4\, m + 1 \\ & m - n + 4 = 0\end{aligned}\right.$ .

Take the sum of these two equations to eliminate the variable $n$ :

$n + (m - n + 4) = 2\, m^2 - 4\, m + 1$ .

Simplify and solve for $m$ :

$2\, m^2 - 5\, m -3 = 0$ .

$(2\, m + 1)\, (m - 3) = 0$ .

There are two possible solutions: $m = -1/2$ and $m = 3$ . For each possible $m$ , substitute back to either of the two equations to find the value of $n$ .

$\displaystyle m = -\frac{1}{2}$ corresponds to $n = \displaystyle \frac{7}{2}$ .
$m = 3$ corresponds to $n = 7$ .

Hence, the two intersections are at $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ and $(3,\, 7)$ , respectively.

<h3>Line Joining Point Q and the Midpoint of Segment AP</h3>

The coordinates of point $A$ and point $P$ each have two components.

For point $A$ , the $x$ -component is $1$ while the $y$ -component is $(-1)$ .
For point $P$ , the $x$ -component is $3$ while the $y$ -component is $7$ .

Let $M$ denote the midpoint of segment $AP$ . The $x$ -component of point $M$ would be $(1 + 3) / 2 = 2$ , the average of the $x$ -components of point $A$ and point $P$ .

Similarly, the $y$ -component of point $M$ would be $((-1) + 7) / 2 = 3$ , the average of the $y\!$ -components of point $A$ and point $P$ .

Hence, the midpoint of segment $AP$ would be at $(2,\, 3)$ .

The slope of the line joining $\displaystyle \left(-\frac{1}{2},\, \frac{7}{2}\right)$ (the coordinates of point $Q$ ) and $(2,\, 3)$ (the midpoint of segment $AP$ ) would be:

$\displaystyle \frac{\text{Change in $y$}}{\text{Change in $x$}} = \frac{3 - (7/2)}{2 - (-1/2)} = \frac{1}{5}$ .

Point $(2,\, 3)$ (the midpoint of segment $AP$ ) is a point on that line. The point-slope form of this line would be:

$\displaystyle \left( y - \frac{7}{2}\right) = \frac{1}{5}\, \left(x - \frac{1}{2} \right)$ .

Rearrange to obtain the slope-intercept form, as well as the standard form of this line:

$\displaystyle y= - \frac{1}{5}\, x + \frac{17}{5}$ .

$x + 5\, y - 17 = 0$ .

7 0

1 year ago

Bill drives and sees a red light. He slows down to a stop. A graph of his velocity over time is shown below. What is his average

Feliz [49]

Answer:

Step-by-step explanation:Velocity at t= 0s is 20m/s and the velocity at t= 10s is 20m/s.

a= 20-20/ 10s

a=0m/s

4 0

1 year ago

Each month Liz pays $35 to her phone company just to use the phone. Each text she sends costs her an additional $0.05. In March

olchik [2.2K]

First let's set up an equation: 35+0.05t.
35 is the phone rate per month plus the number of texts (t) times 0.05 (cost of 1 text)

Now we can find the amount of texts she sent each month.

March
35+0.05t=72.60
-35 -35
0.05t=37.60
------ -------
0.05 0.05
t=752

April
35+0.05t=65.85
-35 -35
0.05t=30.85
------ -------
0.05 0.05
t=617

She sent 752 texts in March and 617 texts in April.

7 0

1 year ago

If ΔABC ≅ ΔFDE, which of the following statements is true?

Fed [463]

Answer:

Option D) ∠B ≅ ∠D

Step-by-step explanation:

we know that

If two triangles are congruent, then the corresponding sides and the corresponding angles are congruent

If ΔABC ≅ ΔFDE

the corresponding angles are

∠A and ∠F

∠B and ∠D

∠C and ∠E

∠A ≅ ∠F

∠B ≅ ∠D

∠C ≅ ∠E

therefore

The statement that is true is ∠B ≅ ∠D

7 0

1 year ago

Read 2 more answers

Round 3,872.3155 to the nearest hundred

zlopas [31]

Answer:

3,872.32

Step-by-step explanation:

the nearest hundredth is 2 numbers on the right of the decimal sign. So you round up from 5 to get answer above.

6 0

2 years ago