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Margaret [11]
3 years ago
15

What is the x value such that y =15 - 3x and y = 0? Please explain how you got this answer.

Mathematics
1 answer:
Mumz [18]3 years ago
8 0

Answer:

X=5

Step-by-step explanation:

Look at the problem | y=15-3x

5×3 is 15.

y=0

15-15=0

Therefore that's the answer.

you're welcome, get better!! :D

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3.5 hours: $10 1/2 hour: $1.25. which is the better deal
andrey2020 [161]
3.5 hours: $10 because if a 1/2 hour equaled 1.25 : 3.5 hours would give you $7.75
8 0
3 years ago
hi im a kid named tyler uzing mi big broters acont to ask this qwestion. wats 46+13? tanks for the help
amm1812

Answer:

59

Step-by-step explanation:

ur welcome

6 0
3 years ago
Area= 18yd squared, length= 6yd<br> width=
GenaCL600 [577]

Answer:

The width is 3 yds

Step-by-step explanation:

Area = length * width

18 = 6*w

Divide each side by 6

18/6 = 6w/6

3=w

The width is 3 yds

7 0
3 years ago
(A journey by bus takes 45 minutes at 60 km per hour. How fast must a car go to do the same
Vlad [161]

Answer:

The car must drive at 108 km/h to cover the same journey.

Step-by-step explanation:

Given that:

Time taken by bus journey = 45 minutes = \frac{45}{60} = 0.75\ hour

Speed of bus = 60 km/h

Distance of the journey = Speed * Time

Distance of the journey = 60 * 0.75 = 45 km

The car needs to cover 45 km in 25 minutes.

25 minutes = \frac{25}{60}\ hour

Distance = Speed * Time

45 = \frac{25}{60}*Speed

Speed = 45*\frac{60}{25}

Speed = 108 km/h

Hence,

The car must drive at 108 km/h to cover the same journey.

7 0
2 years ago
PLEASE HELP WITH GRADE 11 MATH. Make sure you show the formula. Substitute values and show all mathematicall operations! show yo
Degger [83]

3x+y

x

​

 

=−3

=−y+3

​

The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:

\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}

3x+y

3(−y+3)+y

−3y+9+y

−2y

y

​

 

=−3

=−3

=−3

=−12

=6

​

Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:

\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}

x

x

x

​

 

=−y+3

=−(6)+3

=−3

​

The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.

We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.

\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}

3x+y

3(−3)+6

−9+6

−3

​

 

=−3

=

?

−3

=

?

−3

=−3

​

Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:

\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}

7x+10y

−2x+y

​

 

=36

=9

​

In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:

\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}

−2x+y

y

​

 

=9

=2x+9

​

Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:

\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}

7x+10y

7x+10(2x+9)

7x+20x+90

27x+90

3x+10

3x

x

​

 

=36

=36

=36

=36

=4

=−6

=−2

​

Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:

\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}

y

y

y

y

​

 

=2x+9

=2(−2)+9

=−4+9

=5

​

The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.

5 0
2 years ago
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