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Lady_Fox [76]
2 years ago
6

If y ∝ 1∕x and y = –2 when x = 14, find the equation that connects x and y.

Mathematics
1 answer:
crimeas [40]2 years ago
8 0

C. y= -28/x

y=k/x

cross multiply

k= y×x

k = -2×14

k = -28

y = -28/x [ equation connecting x and y]

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A food truck sells sandwiches and
anyanavicka [17]

the answer: is $28 dollars

8 0
2 years ago
What is true about the completely simplified sum of the polynomials 3x2y2 − 2xy5 and −3x2y2 + 3x4y?
svet-max [94.6K]

Answer:

The simplified sum of these polynomials is 3x^4y - 2xy^5

Step-by-step explanation:

In order to find this, we need to remember that we can only add together like terms in this case, there are only two like terms. Both of the first terms end in x^2y^2. So, we add these two together.

3x^2y^2 - 3x^2y^2 = 0

Since they cancel out, we simply just put the other two terms as our answer.

3x^4y - 2xy^5

4 0
2 years ago
Read 2 more answers
2. A parent pledged $0.50 per lap in a walk-thon at school.
Readme [11.4K]
But how many laps did she walk
3 0
2 years ago
If EF bisects CD, CG = 5x-1, GD = 7x-13, EF=6x-4, and GF = 13, find EG (MUST SHOW WORK)
iogann1982 [59]

Answer:

EG = 19

Explanation:

Given that a line segment CD. EF bisects the line CD at point G.

Length of CG = 5x-1

Length of GD = 7x - 13

We know that CG = GD

5x-1 = 7x-13

Solve for x,

2x = 12

x = 6

Now given that,

EF = 6x-4

GF = 13

EF = EG + GF

6x - 4 = EG + 13

EG = 6x - 4 - 13

EG = 6x - 17

put the value of x = 6, in order to find the EG

EG = 6*6 - 17

EG = 19

That's the final answer.

I hope it will help you.

3 0
2 years ago
Express the vector r~ a b c d p r in terms of a~ , b~ , c~ , and d~ , the edges of a parallelogram.
Gre4nikov [31]

Given the vector \vec{r}(a,b,c,d).

Now taking the vertices of parallelogram asA \vec{a},B\vec{b},C\vec{c},D\vec{d}.

As we know to find the edges of parallelogram ABCD,we proceed as follows

\vec{AB}= Position vector of B - Position vector of A

                             \vec{b}-\vec{a}

\vec{BC}= Position vector of C - Position vector of B

                            =\vec{c}-\vec{b}

\vec{CD}= Position vector of D - Position vector of C

                           = \vec{d}-\vec{c}

\vec{DA}= Position vector of A - Position vector of D

                       =\vec{a}-\vec{d}

So, \vec{b}-\vec{a},\vec{c}-\vec{b},\vec{d}-\vec{c},\vec{a}-\vec{d} are edges of parallelogram.

   



                             


                             


6 0
2 years ago
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