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astra-53 [7]
2 years ago
7

I need help, please!! I'll give a brainliest

Mathematics
1 answer:
kvasek [131]2 years ago
3 0

Answer:

metals or metalloids

Step-by-step explanation:

I'm built different i dont know if im right about the question. im just not sure if i understood the question correctly

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It took bob 55min to clean the garage. How many seconds did it take bob.
Setler79 [48]
60x55= 3300 there are 60 seconds in a minute and there are 55 minutes so you simply multiply the seconds by the minutes
5 0
3 years ago
Read 2 more answers
Prove the polynomial identity.
Kaylis [27]

Answer:

Proven

Step-by-step explanation:

We prove the polynomial with factorization. We must first simplify the expression before we can factorize:

(4\cdot{x^2}-2)\cdot{(4\cdot{x^2}-2)}-16\cdot{x^4}

16\cdot{x^4}-8\cdot{x^2}-8\cdot{x^2}+4-16\cdot{x^4}

4-16\cdot{x^2}

4 is the common number and we can take it out of the expression:

4\cdot{(1-4\cdot{x^2})}

We can factorize further with the x term being zero. For this to be true we must have -2x and 2x to cancel out and therefore the expression is proven:

4\cdot{(1-2\cdot{x})}\cdot{(1+2\cdot{x})}

5 0
3 years ago
1 point
sergey [27]

Answer:

the answer is 97 student tickets and 15 adult tickets

Step-by-step explanation:

4 0
2 years ago
Please someone help me solve this i dpnt know the answer​
kozerog [31]
<h2><u>Answer With Explanation:</u></h2>

<u>Firstly, let's start with <XOZ: =55°</u>

We know that <ZOQ is 70° and angles on a line add up to 180° so we do 180-70=110 110 divided by 2 = 55 so the 2 angles (XOZ & XOP are 55)

<u>Secondly, <OMN, <MON & <ONM = All are 60°</u>

These 2 angles are joined to create an equilateral triangle which always adds up to 180°

So, there are 3 points to this triangle, therefore we divide 180 by 3 which is 60. The angles are 60°

<u>Thirdly, <QON: =55°</u>

This angle lies on the line XON which needs to add up to 180°

As we worked out before, <XOZ was 55°

So, <ZOQ was already given as 70°

We then do 55+70=125 then 180-125=55°

<QON is 55°

(I'm only in Grade 9 LOL)

4 0
2 years ago
What is the mode and range of 12,11,14,16,14,12,16,14
Nina [5.8K]
I think the mode is 9.875
6 0
3 years ago
Read 2 more answers
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