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3 years ago

Please help- as soon as possible- Multiply (a+2) (a-4)=

Mathematics

1 answer:

Tresset [83]3 years ago

8 0

Answer:

a2+2a−8

Step-by-step explanation:

Expand (a+4)(a−2)(a+4)(a-2) using the FOIL Method.

a⋅a+a⋅−2+4a+4⋅−2

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3 years ago

Help with homework plzzzzz!!!!!!!!!!!!!

Zanzabum

I'm not good with intigers, but if you added two negative intigers together wouldn't it still be negative? If you added -1 to -2 you'd get -3. It would always be less than the two original integers.

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3 years ago

Solve the equation −96=3(8x)^(5/3).

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3 years ago

The prices of commodities X,Y,Z are respectively x, y, z, rupees per unit. Mr. A purchases 4 units of Z and sells 3 units of X a

liubo4ka [24]

Answer:

$(x,y,z)=(1477, 1464, 1437)$

Step-by-step explanation:

Consider the selling of the units positive earning and the purchasing of the units negative earning.

Mr. A purchases 4 units of Z and sells 3 units of X and 5 units of Y
Mr.A earns Rs6000

So, the equation would be

$3x + 5y - 4z = 6000$

Mr. B purchases 3 units of Y and sells 2 units of X and 1 units of Z
Mr B neither lose nor gain meaning he has made 0₹

hence,

$2x - 3y + z = 0$

Mr. C purchases 1 units of X and sells 4 units of Y and 6 units of Z
Mr.C earns 13000₹

therefore,

$- x + 4y + 6z = 13000$

Thus our system of equations is

$\begin{cases}3x + 5y - 4z = 6000\\2x - 3y + z = 0\\ - x + 4y + 6z = 13000\end{cases}$

<u>Solving </u><u>the </u><u>system </u><u>of </u><u>equations</u><u>:</u>

we will consider elimination method to solve the system of equations. To do so ,separate the equation in two parts which yields:

$\begin{cases}3x + 5y - 4z = 6000\\2x - 3y + z = 0\end{cases}\\\begin{cases}2x - 3y + z = 0\\ - x + 4y + 6z = 13000\end{cases}$

Now solve the equation accordingly:

$\implies\begin{cases}11x-7y=6000\\-13x+22y=13000\end{cases}$

Solving the equation for x and y yields:

$\implies\begin{cases}x= \dfrac{223000}{151}\\\\y= \dfrac{221000}{151}\end{cases}$

plug in the value of x and y into 2x - 3y + z = 0 and simplify to get z. hence,

$\implies z= \dfrac{217000}{151}$

Therefore,the prices of commodities X,Y,Z are respectively approximately 1477, 1464, 1437

6 0

3 years ago