All categories

3 years ago

A postulate can also be called a(n)

1 answer:

4 0

A postulate can also be called an axiom. Answer:A
The postulate is a statement that is accepted as true and it is used as an argument in scientific investigation. It is a claim to take for granted. Axiom is the same. It is a statement that is assumed to be true without proof. Axioms are the basic statements from which all theorems are proved.

You might be interested in

Rewrite using a single exponent. 53.53

Helga [31]

Answer:

${5}^{6} \: \: or \: \: {25}^{3} = 15625$

Step-by-step explanation:

${5}^{3} \times {5}^{3}$

Here we can solve this in 2 ways ,

Method 1 :

${5}^{3} \times {5}^{3}$

Here the bases are equal. So,

$= > {5}^{3 + 3} = {5}^{6} = 15625$

Method 2 :

${5}^{3} \times {5}^{3}$

Here the exponents are same. So,

$= > {(5 \times 5)}^{3} = {25}^{3} = 15625$

4 0

3 years ago

HELP ASAP (Geometry)

Andrei [34K]

1) Parallel line: $y=-2x-3$

2) Rectangle

3) Perpendicular line: $y = 0.5x + 2.5$

4) x-coordinate: 2.7

5) Distance: $d=\sqrt{(4-3)^2+(7-1)^2}$

6) 3/8

7) Perimeter: 12.4 units

8) Area: 8 square units

9) Two slopes of triangle ABC are opposite reciprocals

10) Perpendicular line: $y-5=-4(x-(-1))$

Step-by-step explanation:

The equation of a line is in the form

$y=mx+q$

where m is the slope and q is the y-intercept.

Two lines are parallel to each other if they have same slope m.

The line given in this problem is

$y=-2x+7$

So its slope is $m=-2$ . Therefore, the only line parallel to this one is the line which have the same slope, which is:

$y=-2x-3$

Since it also has $m=-2$

We can verify that this is a rectangle by checking that the two diagonals are congruent. We have:

- First diagonal: $d_1 = \sqrt{(-3-(-1))^2+(4-(-2))^2}=\sqrt{(-2)^2+(6)^2}=6.32$

- Second diagonal: $d_2 = \sqrt{(1-(-5))^2+(0-2)^2}=\sqrt{6^2+(-2)^2}=6.32$

The diagonals are congruent, so this is a rectangle.

Given points A (0,1) and B (-2,5), the slope of the line is:

$m=\frac{5-1}{-2-0}=-2$

The slope of a line perpendicular to AB is equal to the inverse reciprocal of the slope of AB, so:

$m'=\frac{1}{2}$

And using the slope-intercept for,

$y-y_0 = m(x-x_0)$

Using the point $(x_0,y_0)=(7,1)$ we find:

$y-1=\frac{1}{2}(x-7)$

And re-arranging,

$y-1 = \frac{1}{2}x-\frac{7}{2}\\y=\frac{1}{2}x-\frac{5}{2}\\y=0.5x-2.5$

The endpoints of the segment are X(1,2) and Y(6,7).

We have to divide the sgment into 1/3 and 2/3 parts from X to Y, so for the x-coordinate we get:

$x' = x_0 + \frac{1}{3}(x_1 - x_0) = 1+\frac{1}{3}(6-1)=2.7$

The distance between two points $A(x_A,y_A)$ and $B(x_B,y_B)$ is given by

$d=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$

In this problem, the two points are

E(3,1)

F(4,7)

So the distance is given by

$d=\sqrt{(4-3)^2+(7-1)^2}$

We have:

A(3,4)

B(11,3)

Point C divides the segment into two parts with 3:5 ratio.

The distance between the x-coordinates of A and B is 8 units: this means that the x-coordinate of C falls 3 units to the right of the x-coordinate of A and 5 units to the left of the x-coordinate of B, so overall, the x-coordinate of C falls at

$\frac{3}{3+5}=\frac{3}{8}$

of the distance between A and B.

To find the perimeter, we have to calculate the length of each side:

$d_{EF}=\sqrt{(x_E-x_F)^2+(y_E-y_F)^2}=\sqrt{(-1-2)^2+(6-4)^2}=3.6$

$d_{FG}=\sqrt{(x_G-x_F)^2+(y_G-y_F)^2}=\sqrt{(-1-2)^2+(3-4)^2}=3.2$

$d_{GH}=\sqrt{(x_G-x_H)^2+(y_G-y_H)^2}=\sqrt{(-1-(-3))^2+(3-3)^2}=2$

$d_{EH}=\sqrt{(x_E-x_H)^2+(y_E-y_H)^2}=\sqrt{(-1-(-3))^2+(6-3)^2}=3.6$

So the perimeter is

p = 3.6 + 3.2 + 2 + 3.6 = 12.4

The area of a triangle is

$A=\frac{1}{2}(base)(height)$

For this triangle,

Base = XW

Height = YZ

We calculate the length of the base and of the height:

$Base =XW=\sqrt{(x_X-x_W)^2+(y_X-y_W)^2}=\sqrt{(6-2)^2+(3-(-1))^2}=5.7$

$Height =YZ=\sqrt{(x_Y-x_Z)^2+(y_Y-y_Z)^2}=\sqrt{(7-5)^2+(0-2)^2}=2.8$

So the area is

$A=\frac{1}{2}(XW)(YZ)=\frac{1}{2}(5.7)(2.8)=8$

A triangle is a right triangle when there is one right angle. This means that two sides of the triangle are perpendicular to each other: however, two lines are perpendicular when their slopes are opposite reciprocals. Therefore, this means that the true statement is

"Two slopes of triangle ABC are opposite reciprocals"

10)

The initial line is

$y=\frac{1}{4}x-6$

A line perpendicular to this one must have a slope which is the opposite reciprocal, so

$m'=-4$

Using the slope-intercept form,

$y-y_0 = m'(x-x_0)$

And using the point

$(x_0,y_0)=(-1,5)$

we find:

$y-5=-4(x-(-1))$

Learn more about parallel and perpendicular lines:

brainly.com/question/3414323

brainly.com/question/3569195

#LearnwithBrainly

8 0

3 years ago

Here is some information about a holiday.

maksim [4K]

Answer:

$625.6

Step-by-step explanation:

Information about the holiday:

7 night holiday

$340 per person

8% discount if you book before 31 March

Number of people Naseem booked the holiday for = 2

Date of booking of the holiday = 15 February

Total cost of the holiday per person = cost per person - discount before March 31

= $340 - 8% of $340

= 340 - 8/100 * 340

= 340 - 0.08 * 340

= 340 - 27.2

= $312.8

Total cost of the holiday for 2 persons = 2 × Total cost of the holiday per person

= 2 * $312.8

= $625.6

8 0

3 years ago

How do you do it?I'm lost

Leto [7]

Answer and Step-by-step explanation:

The way you complete this is by taking each points x and y and applying the translation. In this case the translation is x-7 and y+1.

Q's x and y are 1 , 1

R's x and y are 1 , 5

T's x and y are 5 , 1

S's x and y are 5 , 5

Now what you do is use the translation on all of these values.

Q's x and y turns into -6 , 2 (subtracted 7 to the 1 and added 1 to the 1)

R's x and y turns into -6 , 6

T's x and y are -2 , 2

S's x and y are -2 , 6

You can then plot the points on the graph and connect the lines to complete the reflection.

Hope this helps ! !

4 0

3 years ago

Anyone wanna have a convo? If you keep the conversation up I'll give brainliest :)

Dmitry [639]

Answer:

Step-by-step explanation:

uhm sure

8 0

2 years ago

Read 2 more answers