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2 years ago

Select the correct answer. What is the decimal form of 17%?A. 17B. 1. 7C. 0. 17D. 0. 17

Mathematics

2 answers:

WITCHER [35]2 years ago

8 0

Answer:

Step-by-step explanation:

You would move the decimal back only 2 places.

kodGreya [7K]2 years ago

8 0

Answer:

0.17

Step-by-step explanation:

All you do is divide by 100.

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Will give brainlist if you can help!

kkurt [141]

Answer: 12:5

Step-by-step explanation:

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3 years ago

What are the end points of the triangle called? How are they labeled?

horrorfan [7]

For this case we have that by definition:

A triangle is defined by three lines that are called sides, or by three points called vertices.

We know that:

The vertices of a triangle are labeled with capital letters.
The sides of a triangle are written with lowercase letters.
The angles of a triangle are written similarly to the vertices.

Answer:

The end points of the triangle are called vertices.

They are labeled in capital letters.

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2 years ago

Solve the system of equations with substitution or elimination method. please help 2x+y=4 and 6x+7y=12

GuDViN [60]

Answer:

x=2-y/2

y=12/7-6x/7

Step-by-step explanation:

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2 years ago

Read 2 more answers

Using Newton's Method, what is the formula for determining the nth term in a sequence of approximations to a solution of a polyn

lorasvet [3.4K]

Step-by-step explanation:

in newton rapson method

$x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}$

Let $n=n-1$

$x_{(n-1)+1}=x_{n-1}-\frac{f\left(x_{n-1}\right)}{f^{\prime}\left(x_{n-1}\right)}$

$x_{n}=x_{n-1}-\frac{f\left(x_{n-1}\right)}{f^{\prime}\left(x_{n-1}\right)}$

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2 years ago

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago