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Pavel [41]
3 years ago
14

# Express the equations in perpendicular form .

Mathematics
1 answer:
mixer [17]3 years ago
4 0

The solution is in attachment.

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Please help if you can!!!!
storchak [24]

Answer:Well, I don't know what you got so I can't tell you if it is right.

If it works in both equations, it depends of whether your equations are set up correctly.

Here is how I would do this problem.

Let x = no. of hot dogs,y = number of sodas.

First equation is just about the number of things.

x + y = 15

Second equation is about the cost of things.

1.5 x + .75 y = 18

solve x+y = 15 for y  y = 15-x    substitute into second equation

1.5x + .75(15 - x) = 18    

You should get the correct answer for number of hot dogs if you solve this correctly.  Put your answer in the x + y =15 equation to get y.  Then put both x and y into the cost equation and check your answer.

Hope this helps.

Step-by-step explanation:

5 0
3 years ago
I don't understand word problems can someone please answer it for me and I need it ASAP.
maksim [4K]

Answer:

Inequality: 3 + 1.2c

What you'd put on graph: 1 ≥ 13.50

5 0
4 years ago
Please help me with average
Anvisha [2.4K]

Answer:

2x+1

Step-by-step explanation:

(3x-5)+(x+2)+(2x+6)=6x+3

6x+3/3=2x+1

8 0
3 years ago
Read 2 more answers
Need help question 11
Helga [31]
I am sorry i dont understand

5 0
3 years ago
Let V=ℝ2 and let H be the subset of V of all points on the line 4x+3y=12. Is H a subspace of the vector space V?
sergejj [24]

Answer:

No, it isn't.

Step-by-step explanation:

We have V=IR^{2} and let H be the subset of V of all points on the line

4x+3y=12

We need to find if H is a subspace of the vector space V.

In IR^{2} all the possibilities for own subspace of the vector space IR^{2} are :

  • IR^{2} itself.
  • The vector 0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]
  • All lines in IR^{2} that passes through the origin  (  0_{IR^{2}}=\left[\begin{array}{c}0&0\end{array}\right]  )

We know that H is the subset of IR^{2} of all points on the line 4x+3y=12

If we look at the equation, the point \left[\begin{array}{c}0&0\end{array}\right] doesn't verify it because :

4x+3y=12\\4(0)+3(0)=12\\0=12

Which is an absurd. Therefore, H doesn't contain the origin (and H is a line in IR^{2}). Finally, it can't be a vector space of V=IR^{2}

8 0
3 years ago
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