The corresponding lengths of two cuboids are 12 and 3. What is the ratio of their volumes?

The vertices of a right triangle are (0, 0), (1, 0), and (0, 1).

Alchen [17]

Answer:

(3,3)

Step-by-step explanation:

Given

$Vertices: (0,0)\ (1,0)\ (0,1)$

$Scale\ Factor = 3$

Required

Determine which can't be any of the new vertices

First, we need to determine the new vertices:

$New\ Vertex = Scale\ Factor * Old\ Vertex$

For (0,0):

$New\ Vertex = 3 * (0,0)$

$New\ Vertex = (3 * 0,3 * 0)$

$New\ Vertex = (0,0)$

For (1,0):

$New\ Vertex = 3 * (1,0)$

$New\ Vertex = (3 * 1,3 * 0)$

$New\ Vertex = (3,0)$

For (0,1):

$New\ Vertex = 3 * (0,1)$

$New\ Vertex = (3 * 0,3 * 1)$

$New\ Vertex = (0,3)$

Comparing the calculated new vertices to the list of given options; (3,3) can't be any of the new vertices of the new triangle

5 0

3 years ago

Question 27 Find the length of the hypotenuse. Leave answer in simplest radical form.

skelet666 [1.2K]

Use SOHCAHTOA. Since we are looking for the hypotenuse, you must use sine or cosine. Since, we are given the adjacent side, use cosine.

cosx=a/h

cos45=8/h

√2/2=8/h

*Cross multiply*

h√2=16

*Divide both sides by √2*

h=16/√2

*Multiply by √2/√2*

h=(16√2)/2

*Simplify the fraction*

h=8√2

Hope this helps!!

6 0

3 years ago

How do I simply this, thanks!

stepan [7]

(-a^3b^2*-a^-2b^-3)^-2/2a^2b^-3 = a^4b^9/2a^8b^4 =b^5/2a^4 so your answer is b^5/2a^4

7 0

3 years ago

Express each of the following quadratic functions in the form of f(x) = a (x - h)²+ k.Then,state the minimum or maximum value,ax

Ivan

Given:

The function is:

$f(x)=-2x^2+7x+4$

To find:

Express the quadratic equation in the form of $f(x)=a(x-h)^2+k$ , then state the minimum or maximum value,axis of symmetry and minimum or maximum point.

Solution:

The vertex form of a quadratic function is:

$f(x)=a(x-h)^2+k$ ...(i)

Where, a is a constant, (h,k) is the vertex and x=h is the axis of symmetry.

We have,

$f(x)=-2x^2+7x+4$

It can be written as:

$f(x)=-2\left(x^2-3.5x\right)+4$

Adding and subtracting square of half of coefficient of x inside the parenthesis, we get

$f(x)=-2\left(x^2-3.5x+(\dfrac{3.5}{2})^2-(\dfrac{3.5}{2})^2\right)+4$

$f(x)=-2\left(x^2-3.5x+(1.75)^2\right)-2\left(-(1.75)^2\right)+4$

$f(x)=-2\left(x-1.75\right)^2+2(3.0625)+4$

$f(x)=-2\left(x-1.75\right)^2+6.125+4$

$f(x)=-2\left(x-1.75\right)^2+10.125$ ...(ii)

On comparing (i) and (ii), we get

$a=-2,h=1.75,k=10.125$

Here, a is negative, the given function represents a downward parabola and its vertex is the point of maxima.

Maximum value = 10.125

Axis of symmetry : $x=1.75$

Maximum point = (1.75,10.125)

Therefore, the vertex form of the given function is $f(x)=-2\left(x-1.75\right)^2+10.125$ , the maximum value is 10.125, the axis of symmetry is $x=1.75$ and the maximum point is (1.75,10.125).

8 0

3 years ago

6b-9=-87 what is the answer

maw [93]

Answer:

Answer is below

Step-by-step explanation:

Add 9 to both sides.

$6b-9+9=-87+9\\6b=-78$