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3 years ago

What are three ratios equvilent to 8:20

Mathematics

1 answer:

Genrish500 [490]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

Multiplying or dividing each part of the ratio gives an equivalent ratio

Given

8 : 20 ← divide both parts by 2

= 4 : 10 ← divide both parts by 2

= 2 : 5 ← ratio in simplest form

Multiply both parts by 2

8 : 20 = 16 : 40

4 : 10, 2 : 5, 16: 40 ← are possible equivalent ratios

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Mia wants to buy 3 notebooks for $1.29 each. What is the expression for the total cost?

ololo11 [35]

Step-by-step explanation:

Mia want to buy 3 notebook for $1.29 each ,

then total cost =3×1.29

hence, 3×29

1\6+1/12+2/6=2+1+4/12

=7/12

8 0

3 years ago

Tryna figure this out

kaheart [24]

Answer:

9pi approximately 28

Step-by-step explanation:

we have the formula - pi r^2

which we know the diameter is 6 m and the surface is a semi circle

the radius is 3 so then we have pi 3^2

which becomes 9pi which is about 28.2

rounding to 28

hope this helps!!

7 0

3 years ago

In XYZ, mX=90 and mY=30.In TUV,mU= 30 and mV=60. Which is true about the two triangles? XYZ=TUV, XYZ= VUT,No congruency statemen

Flauer [41]

Answer:

No congruency statement can be made because the side length are unknown.

Step-by-step explanation:

The sum of the measures of all interior angles in a triangle is always 180°.

In triangle XYZ,

$m\angle X=90^{\circ}\\ \\m\angle Y=30^{\circ},$

then

$m\angle X+m\angle Y+m\angle Z=180^{\circ}\Rightarrow m\angle Z=180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$

In triangle TUV,

$m\angle U=30^{\circ}\\\\m\angle V=60^{\circ},$

then

$m\angle T+m\angle U+m\angle V=180^{\circ}\Rightarrow m\angle T=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}$

We have two triangles with congruent angles but we have no information about side lengths. Therefore, can not be made because the side length are unknown. Correct option is last option.

6 0

3 years ago

Read 2 more answers

Use the Counting Principle to find the probability. Choosing the 7 winning lottery numbers when the numbers are chosen at random

frozen [14]

Answer: 120

<u>Step-by-step explanation:</u>

Since the order of the numbers doesn't matter we can use the formula:

$\dfrac{n!}{r!(n-r)!}\quad \text{where n is the quantity of items and r is the quantity chosen}\\\\\text{In this problem, there are 10 numbers (n = 10) and 7 to be chosen (r = 7)}\\\\_{10}C_{7}=\dfrac{10!}{7!(10-7)!}\\\\.\qquad=\dfrac{10!}{7!3!}\\\\.\qquad=\dfrac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}\\\\.\qquad=10\times 3\times 4\\\\.\qquad=120$

7 0

3 years ago

Read 2 more answers

(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

$\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]$

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0

3 years ago