2x + y = -2x + y = 5

The x-coordinate of the solution to the system shown is _____.

Answers

Answer 1
Answer: The x-coordinate of the solution to the system shown is -7.

Proof:
From the 2nd equation, equate in terms of y:
y = 5 - x
Then substitute it to the first equation:
2x + 5 - x = -2
2x - x = -2 -5
x = -7

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Evaluate. 8.76 + (–3.05)
a. –11.81
b. –5.71
c. 5.71
d. 11.81

Answers

8.76 + (–3.05)
= 8.76 – 3.05
= 5.71
The answer is c.

(12.8)2 - (30+0.375)

Answers

Answer:

-4.775

Step-by-step explanation:

Question 3.3. Find the LCD of these fractions.4/x^2y, 20/xy


(Points : 1)

A. xy
b. xy^2
c.x^2y
d. x^2y^2

help please? :)

Answers

did you ever get an answer?

Tickets for a school play cost $4 for adults and $2 per students. At the endof the play,the school sold 105 tickets and collected $360. -write a linear system that models the situation.
-find the number of adult and students tickets sold.

Answers

4x+2y=360
x+y=  105 => 4x+4y= 4*105= 420
                    4x+2y =360
-----------------------------------------
                    4y-2y = 420-360= 60
                     2y   = 60    => y= 60:2=30 => x= 105-30=75
Verify 4*75+2*30= 300 + 60 = 360
First set your variables: let x be the number of adult tickets and let y be the number of student tickets. Now when you write 4x it will say how much money came from adult tickets, and 2y will do the same for students. 4x+2y= the amount of money collected.
4x+2y=360
Solve the above equation for x and y and that will give you the number of tickets sold.

What is true about the domain and range of the function?The graph of the function f(x) = (x +2)(x + 6) is shown
below.
у
6
The domain is all real numbers, and the range is all
real numbers greater than or equal to 4.
O The domain is all real numbers greater than or equal to
4, and the range is all real numbers.
O The domain is all real numbers such that -65x3-2,
and the range is all real numbers greater than or equal
to-4.
The domain is all real numbers greater than or equal to
4, and the range is all real numbers such that -63x3
-2.
4
N
-6 -4 +2
4
6
х
N
-2-
-4
6t

Answers

Answer:

The domain is all real numbers and the range is all real numbers greater than or equal to -4

Step-by-step explanation:

Ted throws an object into the air with an initial vertical velocity of 54 feet per second from a platform that is 40 feet above the ground. About how long will it take the object to hit the ground?

Answers

Answer:

The height of the object as a function of time is given by:

h(t) = -16t^2+vt+h       .....[1]

where,

h(t) is the height of the object t second after it is thrown.

v is the initial velocity

h is the initial height.

As per the statement:

Ted throws an object into the air with an initial vertical velocity of 54 feet per second from a platform that is 40 feet above the ground

⇒v = 54 ft/sec and h = 40 feet above the ground.

Substitute in [1] we have;

h(t) = -16t^2+54t+40

we have to find how long will it take the object to hit the ground.

Substitute h(t) = 0

then;

-16t^2+54t+40 = 0

-8t^2+27t+20=0

Factorize this equation:

-8t^2+32t-5t+20=0

-8t(t-4)-5(t-4)=0

(t-4)(-8t-5)=0

By zero product property we have;

t-4 = 0 and -8t-5 = 0

⇒t = 4 and -8t = 5

⇒ t= 4 and t = - 0.625

Since, time cannot be in negative;

⇒ t= 4 sec

therefore, 4 seconds  will it take the object to hit the ground


PERFECT application of the general equation of things
that are moving up and down in gravity:

                               H(t) = H₀ + v₀t + 1/2 g t²

t  =  time after the toss
H(t)  =  height at any time after the toss
H₀  =  height from which it's tossed
v₀ =  initial speed
g  =  acceleration of gravity

I'm going to call 'UP' = the positive direction .
So ...

H₀ = +40 ft
v₀  =  +54 ft/sec
g  =  -32 ft/sec²

When the object hits the ground, H(t) = 0 .
So you have ...

   0  =  40  +  54 t  -  16 t²

Let me re-write that for you, to make it a
little easier and more familiar:

                                        - 16 t²  +  54 t  +  40  =  0       

Divide each side by -2 :      8 t²   -  27 t  -  20  =  0 

Now go to your toolbox, pull out your quadratic formula,
solve that quadratic equation, and you'll get the usual two
solutions. 

The positive one is the one you want. 

The negative one refers to time before the toss, and may be discarded.