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Elodia [21]
3 years ago
6

Factorization x^2+2x+5​

Mathematics
1 answer:
anyanavicka [17]3 years ago
7 0

Answer:

x^ 2 + 2 x + 5

Step-by-step explanation:

The expression is not factorable with rational numbers.

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Please give me the answer to number 2 you don’t need to explain it just give me the answer
Yakvenalex [24]

Hello from MrBillDoesMath!

Answer:   Perimeter = 17x + 4

Discussion:

(So you are trying to sneak #2 out of me too..... );

Perimeter = sum of lengths of sides. Starting at the bottom of the figure and proceeding counterclockwise.

Perimeter =  (3x +2) + (3x+2) + (4x-1) + (4x-1) + (3x + 2) .

Separate and combine x's and constants:

Perimeter = (3x + 3x+4x+4x+3x) + (2 + 2 -1 -1 + 2) =

                = 17x + 4

You answer has the wrong constant value ( you wrote 17x + 1. The constant should be 4)


Thank you,

MrB

8 0
3 years ago
How do i Write (3x^3)^2 without exponents
Murljashka [212]
The answer is there I explained the process.

8 0
3 years ago
I tried using the standard deviation formula but still got the wrong answer. How would I solve this?
nadezda [96]

Answers:

First box = 864

Second box = 3136

=================================================

Explanation:

The weird looking E symbol is the greek uppercase letter sigma which stands for "sum". You'll be adding everything in the x^2 column to find that 64+144+256+400 = 864, which is already done for you in the table (as the 864 is at the bottom of the second column, in the "total" row).

In short, you just copy the 864 into the first box.

-------------

For the second box, you'll square the 56 to get 3136

This is because the table shows that adding up all the x values leads to 56

(8+12+16+20 = 56)

So that means \sum x = 56 and \left(\sum x\right)^2 = (56)^2 = 3136

7 0
3 years ago
Read 2 more answers
I will give brainliest to the first one that does it correctly
Effectus [21]

Answer:

180-138

=42, is the answer

8 0
3 years ago
Cos4theta+cos2theta/ cos4theta-cos2theta= _____
vovangra [49]

\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}

\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )

8 0
3 years ago
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