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4 years ago

Solve for r. 12 = r - (34 - 2) Show your work!

Mathematics

2 answers:

kaheart [24]4 years ago

7 0

Answer:

r = 44

Step-by-step explanation:

We need to find out the value of r that makes true that equation. For that, we need to leave the r alone in one side of the equation and get all the numbers together on the other side.

12 = r - (34 - 2)

First, let's resolve the calculation inside the parenthesis:

34 - 2 = 32

So,

12 = r - 32

Now let's add 32 on both sides of the equation:

12 + 32 = r - 32 + 32

44 = r

djverab [1.8K]4 years ago

3 0

The answer would be -20

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4 years ago

Read 2 more answers

Use the following matrices, A, B, C and D to perform each operation.

Vinvika [58]

Step-by-step explanation:

$A=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]$

$B=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]$

$C=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]$

$D=\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]$

$1.\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]+\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]=\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&2\\11&7\end{array}\right]$

$2.\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]$

$3.\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]$

$4.\\C\cdot D=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\cdot\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\=\left[\begin{array}{ccc}7&-8&-6\\14&13&-8\end{array}\right]$

$5.\\2D+3C\\\text{This operation can't be performed because the matrices}\\\text{ are of different dimensions.}$

6 0

4 years ago

Sequences and series

Vinvika [58]

Answer:

Step-by-step explanation:

$a_{n}=\frac{(n+3)!}{(n+3}=\frac{(n+3)(n+2)!}{n+3}=(n+2)!\\a_{1}=(1+2)!=3!=3*2*1=6\\a_{2}=(2+2)!=4!=4*3*2*1=24\\a_{3}=(3+2)=5!=5*4!=5*24=120\\a_{4}=(4=2)!=6!=6*5!=120*6=720\\a_{5}=(5+2)=7!=7*6!=720*7=5040$