Answer:
(5,5)
Step-by-step explanation:
Answer:
The quotient is 3x - 11 + 60/(x + 5) ⇒ 2nd answer
Step-by-step explanation:
* We will use the long division to solve the problem
- The dividend is 3x² + 4x + 5
- The divisor is x + 5
- The quotient is the answer of the division
- If the divisor not a factor of a dividend, the quotient has
a remainder
* Lets solve the problem
- At first divide the first term in the dividend by the first term in
the divisor
∵ 3x² ÷ x = 3x
- Multiply the divisor by 3x
∴ 3x (x + 5) = 3x² + 15x
-Subtract this expression from the dividend
∴ 3x² + 4x + 5 - (3x² + 15x) = 3x² + 4x + 5 - 3x² - 15x = -11x + 5
- Divide the first term -11x in the new dividend by the first
term x in the divisor
∴ -11x ÷ x = -11
- Multiply the divisor by -11
∴ -11(x + 5) = -11x - 55
-Subtract this expression from the new dividend
∴ -11x + 5 - (-11x - 55) = -11x + 5 + 11x + 55 = 60
∴ The quotient is 3x - 11 with remainder = 60
* The quotient is 3x - 11 + 60/(x + 5)
Negative, Nonlinear,
I hope this helped:)
<span>0.002 x 0.003 = 0.000006
the zeroes are multiplied by the power of 10 which is 1/10 in particular.
For example.
The product of a whole number and a decimal number less than 1 will be greater than the whole number multiplied into. For this theorem to be proven. Let us state the mathematical expression into numbers such that </span><span><span>
1. </span> N x 0.1 = N/0.1 < N</span> <span><span>
2. </span> 1 x 0.5 = 0.5 </span><span><span>
3. </span> 2 x 0.1 = 0.2</span> <span><span>
4. </span> 100 x 0.55 = 55</span><span> </span>
<span>These three examples and stances then suggest the claim that the product is not equal to the whole number used in the equation.<span>
</span></span>
Answer:
![\left[\begin{array}{ccc}32&-55\\3&39\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D32%26-55%5C%5C3%2639%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
![\left[\begin{array}{ccc}0&-15\\3&15\end{array}\right] +\left[\begin{array}{ccc}32&-40\\0&24\end{array}\right] =\left[\begin{array}{ccc}0+32&-15+(-40)\\3+0&15+24\end{array}\right] = \left[\begin{array}{ccc}32&-55\\3&39\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-15%5C%5C3%2615%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D32%26-40%5C%5C0%2624%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%2B32%26-15%2B%28-40%29%5C%5C3%2B0%2615%2B24%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D32%26-55%5C%5C3%2639%5Cend%7Barray%7D%5Cright%5D)
