Please i am dying this is A REALLY HARD QUESTION! 1+4=??!?!?!??

Mathematics

2 answers:

lukranit [14]3 years ago

8 0

Use a calculator its 5

Keith_Richards [23]3 years ago

4 0

The answer that would be 5!

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yesterday Josh sold some boxes of greeting cards. today he sold seven boxes .if he sold 25 boxes in all , how many did he sell

Galina-37 [17]

If he sold 25 boxes in all and he sold 7 today then he would of sold 25 - 7 boxes yesterday. Josh sold 18 boxes yesterday.

Today = 7

Yesterday = 18

Total = 25

7 + x = 25, when x = Yesterdays sold boxes.

Hope this helps :)

7 0

3 years ago

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Enter the range of data 37,45,20,34,28,80

Montano1993 [528]

Answer:

Step-by-step explanation:

To find the range, you subtract the largest number and the smallest number.

Largest number: 80

Smallest number: 20

80-20=60

8 0

3 years ago

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Find the solution to this system of equations <br> 3x+2y+3z=3 4x-5y+7z=1 2x+3y-2z=6 <br> x=? Y=? Z=?

Ede4ka [16]

Answer:

The solution is $x=2,\ y=0,\ z=-1.$

Step-by-step explanation:

You are given the system of three equations:

$\left\{\begin{array}{l}3x+2y+3z=3\\4x-5y+7z=1\\2x+3y-2z=6\end{array}\right.$

Multiply the first equation by 4, the second equation by 3 and subtract them. Then multiply the third equation by 2 and subtract it from the second equation:

$\left\{\begin{array}{l}3x+2y+3z=3\\4(3x+2y+3z)-3(4x-5y+7z)=4\cdot 3-3\cdot 1\\4x-5y+7z-2(2x+3y-2z)=1-2\cdot 6\end{array}\right.\Rightarrow \\\\\left\{\begin{array}{l}3x+2y+3z=3\\12x+8y+12z-12x+15y-21z=12-3\\4x-5y+7z-4x-6y+4z=1-12\end{array}\right.$

So,

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\-11y+11z=-11\end{array}\right.\Rightarrow \left\{\begin{array}{l}3x+2y+3z=3\\23y-9z=9\\y-z=1\end{array}\right.$

Multiply the third equation by 23 and subtract it from the second equation:

$\left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23(y-z)=9-23\cdot 1\end{array}\right.\Rightarrow \left\{\begin{array}{rl}3x+2y+3z=3\\23y-9z=9\\23y-9z-23y+23z=9-23 \end{array}\right.$