If y=x+2 were changed to y=x-1 how would the graph of the new function compare with the first one

If two lines are parallel, corresponding angles are _________.

jolli1 [7]

B. congruent, im pretty sure

6 0

3 years ago

Read 2 more answers

Gina Wilson unit 3: homework 6: Slope intercept form and standard form HELP

Goshia [24]

Answer:

1.) 2x-y=-7 --> y=2x + 7

2.) 5x+3y=12 --> y= -5/3x + 4

3.) x-y=1 --> y= x- 1

4.) x-3y=-15 --> y=1/3x + 5

5.) 8x-10y=20 --> y=4/5x - 2

6.) 14x+6y=36 --> y=-7/3x+6

Step-by-step explanation:

Hope this helped.

8 0

3 years ago

Find the diameter of a circle with an area<br> of 415.48 square inches.

lozanna [386]

D= 23 in.

hope its what you’re looking for :)

8 0

3 years ago

Find AM in the parallelogram

n200080 [17]

Answer:

AM = 6

Step-by-step explanation:

Using the property of a parallelogram

• The diagonals bisect each other

MO is a diagonal, hence

AM = AO = 6

5 0

3 years ago

given the recursive formula for a geometric sequence find the common ratio the 8th term and the explicit formula.did I set these

lesya [120]

Answer:

Step-by-step explanation:

1)Since we know that recursive formula of the geometric sequence is

$a_{n}=a_{n-1}*r$

so comparing it with the given recursive formula $a_{n}=a_{n-1}*-4$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-2*(-4)^{7} =32768.$

Explicit Formula = $-2*(-4)^{n-1}$

2) Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=-4*(-2)^{7} =512.$

Explicit Formula = $-4*(-2)^{n-1}$

3)Comparing the given recursive formula $a_{n}=a_{n-1}*3$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =3

8th term= $a_{1}*(r)^{n-1}=-1*(3)^{7} =-2187.$

Explicit Formula = $-1*(3)^{n-1}$

4)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=3*(-4)^{7} =-49152.$

Explicit Formula = $3*(-4)^{n-1}$

5)Comparing the given recursive formula $a_{n}=a_{n-1}*-4$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-4

8th term= $a_{1}*(r)^{n-1}=-4*(-4)^{7} =65536.$

Explicit Formula = $-4*(-4)^{n-1}$

6)Comparing the given recursive formula $a_{n}=a_{n-1}*-2$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-2

8th term= $a_{1}*(r)^{n-1}=3*(-2)^{7} =-384.$

Explicit Formula = $3*(-2)^{n-1}$

7)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=4*(-5)^{7} =-312500.$

Explicit Formula = $4*(-5)^{n-1}$

8)Comparing the given recursive formula $a_{n}=a_{n-1}*-5$

with standard recursive formula $a_{n}=a_{n-1}*r$

we get common ratio =-5

8th term= $a_{1}*(r)^{n-1}=2*(-5)^{7} =-156250.$

Explicit Formula = $2*(-5)^{n-1}$

6 0

3 years ago