All categories

2 years ago

The graph of a linear system is shown.

Mathematics

1 answer:

kondaur [170]2 years ago

7 0

The solutions to a linear system will always be the interception of the graphs.

In this case, the solution is $(2,7)$

You might be interested in

What are 2 ways to solve the problem 9×1×5 = ?

aliina [53]

2ways you can solve that is by multiplying 9 by 1 then by 5 or 9by 5 then by 1

6 0

2 years ago

When will the ball hit the ground

Likurg_2 [28]

Sorry but you didnt add a picture nor description. If you can, please re-ask this question with the following:

Details to your question

A picture or drawing

Thanks - Madilyn.

5 0

2 years ago

Read 2 more answers

The capacities of four barrels are 75 L , 63 L , 95 L , and 46 L . Plot the values on the vertical number line.

ZanzabumX [31]

It B yo jrjshd. Sudhshsbd she have shav

4 0

2 years ago

30 POINTS,NEED HELP ASAP !!!

zaharov [31]

Answer: OPTION A.

Step-by-step explanation:

You can observe that in the figure CDEF the vertices are:

$C(-2,-1),\ D(-2,0),\ E(2,2)\ and\ F(2,1)$

And in the figure C'D'E'F' the vertices are:

$C'(-8,-4),\ D'(-8,0),\ E'(8,8)\ and\ F'(8,4)$

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:

For C'(-8,-4) and C(-2,-1):

$\frac{-8}{-2}=4\\\\\frac{-4}{-1}=4$

Let's choose another vertex. For E'(8,8) and E(2,2):

$\frac{8}{2}=4\\\\\frac{8}{2}=4$

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.

Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:

$(x, y)$ → $(4x, 4y)$

7 0

2 years ago

Read 2 more answers

Hi can anyone help me with this ..

aleksley [76]

Answer:

n = π/6, π/4, 3π/4, 5π/6

Step-by-step explanation:

sin(3n) − sin n = cos(2n)

Use double and triple angle formulas:

(3 sin n − 4 sin³ n) − sin n = 1 − 2 sin² n

2 sin n − 4 sin³ n = 1 − 2 sin² n

4 sin³ n − 2 sin² n − 2 sin n + 1 = 0

Factor by grouping:

2 sin² n (2 sin n − 1) − (2 sin n − 1) = 0

(2 sin² n − 1) (2 sin n − 1) = 0

Solve:

2 sin² n − 1 = 0

2 sin² n = 1

sin² n = 1/2

sin n = √2/2

n = π/4, 3π/4

2 sin n − 1 = 0

2 sin n = 1

sin n = 1/2

n = π/6, 5π/6

4 0

2 years ago