What is p-1/4•3=-5/6

Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match the system of equations with their so

allochka39001 [22]

Answer:

Part 1) solution {(-5,8),(3,0)} ----> $y+12=x^{2}+x$ and $x+y=3$

Part 2) No solutions ----> $y-15=x^{2}+4x$ and $x-y=1$

Part 3) solution [(-2,5),(3,-5)] ---> $y+5=x^{2}-3x$ and $2x+y=1$

Part 4) No solutions ---> $y-6=x^{2}-3x$ and $x+2y=2$

Part 5) solution [(2,3),(8,9)] ---> $y-17=x^{2}-9x$ and $-x+y=1$

Part 6) solution [(-2,3),(7,-6)] ---> $y-15=-x^{2}+4x$ and $x+y=1$

Step-by-step explanation:

Part 1) we have

$y+12=x^{2}+x$ ----> equation A

$x+y=3$ ----> equation B

Solve by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (-5,8) and (3,0)

see the attached figure N 1

Part 2) we have

$y-15=x^{2}+4x$ ----> equation A

$x-y=1$ ----> equation B

Solve by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The system has no solutions, because there is no point of intersection between the two graphs

see the attached figure N 2

Part 3) we have

$y+5=x^{2}-3x$ ----> equation A

$2x+y=1$ ----> equation B

Solve by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (-2,5) and (3,-5)

see the attached figure N 3

Part 4) we have

$y-6=x^{2}-3x$ ----> equation A

$x+2y=2$ ----> equation B

Solve by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The system has no solutions, because there is no point of intersection between the two graphs

see the attached figure N 4

Part 5) we have

$y-17=x^{2}-9x$ ----> equation A

$-x+y=1$ ----> equation B

Solve by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (2,3) and (8,9)

see the attached figure N 5

Part 6) we have

$y-15=-x^{2}+4x$ ----> equation A

$x+y=1$ ----> equation B

Solve by graphing

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (-2,3) and (7,-6)

see the attached figure N 6

5 0

3 years ago

The third grade class collected coins for the week.Class one raised 34.00. Class two raised 45.00 .Class three raised 76.00.How

Troyanec [42]

Answer:

160 well it was 155 but u round up to 160

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Which of the following statements regarding the expansion of (x + y)^n are correct?

kompoz [17]

Answer:

A, B and D are true statements.

Step-by-step explanation:

We are given a binomial expansion

$(x+y)^n=^nC_0x^ny^0+^nC_1x^{n-1}y^1+^nC_2x^{n-2}y^2+...........+^nC_nx^0y^n$

$(x+y)^n=x^n+nx^{n-1}y+^nC_2x^{n-2}y^2+...........nxy^{n-1}+y^n$

Now we will check each option

Option A: The coefficients of $x^n$ and $y^n$ both equal 1.

If we see first and last term of the expansion, This statement is true.

Option B: For any term $x^ay^b$ in the expansion, a + b = n.

Let we take 3rd term of expansion $^nC_2x^{n-2}y^2$

Here, a=n-2 and b=2

If we do a+b = n-2+2=n

a+b=n is true statement.

Option C: For any term x^ay^b in the expansion, a - b = n.

Let we take 3rd term of expansion $^nC_2x^{n-2}y^2$

Here, a=n-2 and b=2

If we do a-b = n-2-2=n-4≠n

a-b=n is false statement.

Option D: The coefficients of x^ay^b and x^by^a are equal.

If we take second term from beginning and last of the expansion.

$\text{Coefficient From beginning } nx^{n-1}y=n$

$\text{From last } nxy^{n-1}=n$

This statement true.

8 0

3 years ago

The difference of 21 and 16

pogonyaev

Answer:

5

Step-by-step explanation:

i think

6 0

3 years ago

1) A rectangular storage container with an open top is to have a volume of 15 m 3 . The length of its base is four times the wid

My name is Ann [436]

9514 1404 393

Answer:

a) V = 4w²h

b) SA = 4w² +10wh

c) SA = 4w² +37.5/w

d) C = 40w² +225/w

Step-by-step explanation:

The relevant formulas are ...

V = LWH

base area = LW

lateral area = H(2(L+W))

__

a) The length is 4 times the width, so the volume is ...

V = (4w)(w)(h)

V = 4w²h

__

b) The total surface area is the sum of the base area and the lateral area:

SA = base area + lateral area

SA = (4w)(w) + 2h(4w +w)

SA = 4w² +10wh

__

c) The volume is 15 m³, so the height in meters in terms of the width in meters is ...

15 = 4w²h

h = 15/(4w²)

Then the surface area is ...

SA = 4w² +10w(15/(4w²))

SA = 4w² +37.5/w

__

d) The equation we have for surface area has one term for base area and a second term for lateral area. We can apply the cost factors to those terms to get the cost of materials:

C = 10(4w²) +6(37.5/w)

C = 40w² +225/w

4 0

3 years ago