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3 years ago

What are two or more geometric figures that are identical in shape although not necessarily the same size?

Mathematics

1 answer:

Andrei [34K]3 years ago

3 0

Unless I am misreading, the question, or don't know all of it, there can be an infinite number of similar shapes. Three are give to you at the bottom of this answer, but there are many more possibilities.

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Does a shape with a right angle always have perpendicular sides? Explain

Aleks [24]

Answer:

Yes

Step-by-step explanation:

A right angle means that the lines will create an L or X or T when they intersect. This is exactly the definition of perpendicular, that the shapes will intersect and create at least one right angle.

3 0

3 years ago

Read 2 more answers

Let D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36} Determine which of the following statements are true and which are false. a)

morpeh [17]

Answer:

$\forall x\in D$ if x is odd then x> 0 is true statement.

$\forall x\in D$ if x is even then x≤0 is false statement.

$\forall x\in D$ If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

$\forall x\in D$ if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

Step-by-step explanation:

Consider the provided information.

D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}

Part (A) $\forall x\in D$ if x is odd then x> 0

Here only even numbers are less than 0 that means the statement is true.

$\forall x\in D$ if x is odd then x> 0 is true statement.

Part (B) $\forall x\in D$ if x is less than 0 then x is even.

Here only even numbers are less than 0 that means the statement is true.

$\forall x\in D$ if x is odd then x> 0 is true statement.

Part (C) $\forall x\in D$ if x is even then x≤0

Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.

$\forall x\in D$ if x is even then x≤0 is false statement.

Part (D) $\forall x\in D$ If the ones digit of x is 2, then the tens digit is 3 or 4.

There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.

$\forall x\in D$ If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

Part (E) $\forall x\in D$ if the ones digit of x is 6, then the tens digit is 1 or 2.

Numbers having ones digit 6 are: 16, 26 and 36

Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.

$\forall x\in D$ if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

8 0

3 years ago

A rectangular field is two times as long as it is wide. If the perimeter of the field is 450 feet, what are the dimensions of th

lys-0071 [83]

Hi There,

----------------------------------------

We Know:

Total Perimeter = 450 feet

Length is 2 times the width.

----------------------------------------

Solution:

Length = 150

Width = 75

75 x 2 = 150

150 + 150 + 75 + 75 = 450

----------------------------------------

Answer:

Length = 150 feet.

Width = 75 feet.

----------------------------------------

Hope This Helps :)

5 0

3 years ago

Which side lengths form a right triangle.

belka [17]

Answer: B

Explanation:

Use Pythagorean’s theorem to find it out:

A^2 + B^2 = C^2

8^2 + 15^2 = 17^2

64 + 225 = 289

289 = 289

So these side lengths create a right triangle

3 0

3 years ago

Log based question, i just need help with parts b and c

Tanya [424]

Answer:

b) There is translation 3 units to the right and 1 unit up

c) The domain is {x I x > 3}

The equation of the asymptote is x = 3

Step-by-step explanation:

* Lets revise the rule of the translation

- If the function f(x) translated horizontally to the right

by h units, then the new function g(x) = f(x - h)

- If the function f(x) translated horizontally to the left

by h units, then the new function g(x) = f(x + h)

- If the function f(x) translated vertically up

by k units, then the new function g(x) = f(x) + k

- If the function f(x) translated vertically down

by k units, then the new function g(x) = f(x) – k

* Now lets solve the problem

b) ∵ f(x) = $log_{2} (x-3)+1$

∵ The parent function is $log_{2} x$

∴ x is changed to (x - 3), that means there is a translation 3 units

to the right

∵ We add the parent function by 1, that means there is a translation

1 unit up

* There is translation 3 units to the right and 1 unit up

c) To find the domain of the function, find the values of x which

make the function undefined

∵ $log_{2}(0)$ is undefined

∴ x - 3 can not be 0

∵ x - 3 = 0 ⇒ add 3 to both sides

∴ x = 3

∴ The domain of the function is all real number greater than 3

* The domain is {x I x > 3}

∵ x can not be 3

∴ There is a vertical asymptote, its equation is x = 3

* The equation of the asymptote is x = 3

# Look to the attached graph for more understand for the domain

and the equation of the asymptote

4 0

3 years ago