All categories

2 years ago

In the diagram, EFG= OPQ complete the statement

Mathematics

1 answer:

Igoryamba2 years ago

8 0

2) EF is congruent to OP
3) angle P is congruent to angle F
4) angle G is congruent to angle Q (that weird upside down parentheses, I think that is a Q)
5) measure of angle O equals 110 degrees and it equals the measure of angle E
6)QO equals GE and it is 7km
7) triangle GFE is congruent to triangle QPO

You might be interested in

Please help fast please

prohojiy [21]

70-35 I believe that’s write

8 0

2 years ago

Lines m and p are perpendicular. if the slope of line m is -3 what is the slope of line p

ivann1987 [24]

For two lines to be perpendicular, their slopes must be negative reciprocals of one another, mathematically:

m1*m2=-1, in this case our reference line has a slope of -3 so:

-3m=-1 divide both sides by -3

m=-1/-3

m=1/3

So the perpendicular line will have a slope of 1/3

4 0

3 years ago

Ms Feheley buys 9 bottles of eye of Newt for the same price, and also a new couldron for $9.45. She spent $118.89. How much did

scoundrel [369]

See picture for answer and solution steps.

4 0

3 years ago

Read 2 more answers

Generate a line up for the 9players on a ball team

Setler79 [48]

Answer:

do a 3 by 3

Step-by-step explanation:

three across and three up

3 0

3 years ago

Solve the system by using a matrix equation.<br> --4x - 5y = -5<br> -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