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3 years ago

4/7w = 16 hurry please

Mathematics

1 answer:

Novay_Z [31]3 years ago

5 0

Answer:

Multiply both sides of the equation by

⋅

Simplify

⋅

Step-by-step explanation:

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Directions for constructing the in-center of a triangle?

Marat540 [252]

Answer:

HERE YOU GOOO!!!!!

Step-by-step explanation:

1. locate the incenter by constructing the angle bisectors of at least two angles of the triangle.

2. construct a perpendicular from the incenter to one side of the triangle to locate the exact radius.

3. place compass point at the incenter and measure from the center to the point where the perpendicular crosses the side of the triangle (the radius of the circle).

4. draw the circle.

5 0

3 years ago

Solve each equation. Use models if necessary. Check your solution.

morpeh [17]

so, the right answer is -8.

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3 years ago

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Find the prime factorization of the number 20. If the number is prime, state this

mihalych1998 [28]

Answer:

Some of the prime factors are 1,2,4,5,10,20,-1,-.2,-4,-5,-10,-20, and since 5 is prime you can just start with 5 and 4

Step-by-step explanation:

3 0

3 years ago

Simplify . 4y4 4y8 8y4 8y8

Tanya [424]

Answer:

Step-by-step explanation:

Just did it.

4 0

4 years ago

Solve the system by using a matrix equation. --4x - 5y = -5 -6x - 8y = -2

evablogger [386]

Answer:

Solution : (15, - 11)

Step-by-step explanation:

We want to solve this problem using a matrix, so it would be wise to apply Gaussian elimination. Doing so we can start by writing out the matrix of the coefficients, and the solutions ( - 5 and - 2 ) --- ( 1 )

$\begin{bmatrix}-4&-5&|&-5\\ -6&-8&|&-2\end{bmatrix}$

Now let's begin by canceling the leading coefficient in each row, reaching row echelon form, as we desire --- ( 2 )

Row Echelon Form :

$\begin{pmatrix}1\:&\:\cdots \:&\:b\:\\ 0\:&\ddots \:&\:\vdots \\ 0\:&\:0\:&\:1\end{pmatrix}$

Step # 1 : Swap the first and second matrix rows,

$\begin{pmatrix}-6&-8&-2\\ -4&-5&-5\end{pmatrix}$

Step # 2 : Cancel leading coefficient in row 2 through $R_2\:\leftarrow \:R_2-\frac{2}{3}\cdot \:R_1$ ,

$\begin{pmatrix}-6&-8&-2\\ 0&\frac{1}{3}&-\frac{11}{3}\end{pmatrix}$

Now we can continue canceling the leading coefficient in each row, and finally reach the following matrix.

$\begin{bmatrix}1&0&|&15\\ 0&1&|&-11\end{bmatrix}$

As you can see our solution is x = 15, y = - 11 or (15, - 11).

4 0

3 years ago