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Alla [95]
3 years ago
7

Help me in question 7

Mathematics
2 answers:
andriy [413]3 years ago
8 0
I think it’s either b or c
xenn [34]3 years ago
4 0
it’s b! i did this like 4 yrs ago
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Some questions have the form y = kx, where k is a number. Drag each question to the correct category below. ( X = y/2 ),( y = .1
fomenos

Answer:

<u>Has the form y = kx</u>

( y = 0.11x)

(y = 0.04x)

<u>Can be put into form y = kx</u>

(z/x = 9)

<u>Other</u>

(5 = xy)

( x = y/2)

Step-by-step explanation:

<u>y = kx</u>

( y = 0.11x)

(y = 0.04x)

<u>kx = y</u>

(z/x = 9)

-----------

( x = y/2 )

( x - 5 = y)

(5 = xy)

8 0
3 years ago
*WILL GIVE BRIANLIEST PLEASEEEEEEE HELP ME PLEASE* 06.05 MC)
strojnjashka [21]
D. 39.5 is the likely hood of the students likeing burgers and hotdogs based on the information given 
3 0
3 years ago
Read 2 more answers
Can you help me with 22 a and b plz
BartSMP [9]
For a you would just have to add all the velocity up and divide by 5. For b its -15 the ball is going down.
3 0
3 years ago
By the end of Monday, a grocery distribution center has 240 cases of green beans in its warehouse. Every day, the center deliver
asambeis [7]
The answer is B.
Explanation:

240-18
= 222
222-18
= 204
204-18
= 186
6 0
3 years ago
Read 2 more answers
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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