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Evgesh-ka [11]
3 years ago
5

Which ordered pair is a solution of the linear system: x+2y=-8 and

Mathematics
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

option A

A(-2,-3)

Step-by-step explanation:

make y subject of formula

x+2y=-8

2y=-8-x

y=(-8-x)/2

y=-4-x/2

when x=-2

y=-4-(-2)/2

=-4-(-1)

=-4+1

=-3

(-2,-3) is an ordered pair

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Aleksandr [31]
5.76 is the correct answer to your problem
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4 years ago
LET x=0 x-2y=4 what is the value of y?
Amanda [17]

Answer:

When x = 0, then y = -2

When y = 0, then x = 4

Step-by-step explanation:

We are given with the following equation;

x - 2y = 4

Now, we have to find the respective values of x and y for each value of x = 0 and y = 0.

Firstly, putting the value of x = 0;

x - 2y = 4

0 - 2y = 4

-2y = 4

y = \frac{4}{-2}  = -2

This means that when x = 0, then y = -2.

Similarly, putting the value of y = 0;

x - 2y = 4

x-(2\times 0)=4

x - 0 = 4

x = 4

This means that when y = 0, then x = 4.

4 0
3 years ago
Adrian has $190 to spend on his little brother's birthday party. The clown is going to
grandymaker [24]

Answer:

120 + 3.5x≤ 190

Step-by-step explanation:

Add up all the costs

80+ 10+10+20 + 3.50x where x is the number of cupcakes

Combine like terms

120 + 3.5x

This must be less than or equal to 190

120 + 3.5x≤ 190

8 0
3 years ago
Please help me:<br><br>What's the domain?<br><br>What's the range?<br><br>Is it a function?​
VladimirAG [237]

Answer:

\displaystyle Yes \\ Range: Set-Builder\:Notation → [f(x)|-2 ≤ f(x) ≤ 4] \\ Interval\:Notation → [-2, 4] \\ \\ Domain: Set-Builder\:Notation → [x|0 ≤ x ≤ 7] \\ Interval\:Notation → [0, 7]

Step-by-step explanation:

Just by looking at the graph vertically and horisontally, you can tell what the range and domain is, depending on whether the segments are <em>closed</em> or <em>opened</em>.

* This is a function because it passes the <em>vertical</em><em> </em><em>line test</em>.

** This is kind of like a sine wave.

I am joyous to assist you anytime.

6 0
3 years ago
Read 2 more answers
Please prove this........​
Crazy boy [7]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    →     C = π - (A + B)

                                    → sin C = sin(π - (A + B))       cos C = sin(π - (A + B))

                                    → sin C = sin (A + B)              cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

<u>Proof LHS → RHS</u>

LHS:                        (sin 2A + sin 2B) + sin 2C

\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C

\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C

\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C

\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)

\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]

\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B

\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C    \checkmark

7 0
4 years ago
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