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liubo4ka [24]
3 years ago
10

‼️ WILL LIST BRAINLIEST ‼️Children learn personal ____ and life skills from their families

Mathematics
2 answers:
ad-work [718]3 years ago
6 0

Answer:

D

Step-by-step explanation:

Lostsunrise [7]3 years ago
4 0

Answer:

Emotions

Hope this helps

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Increase 200 by 17% <br><br><br><br> please help
Ugo [173]
The answer for this question is 234
3 0
3 years ago
Read 2 more answers
What is the answer to this equation <img src="https://tex.z-dn.net/?f=10%5Csqrt%7Bx%7D%2020" id="TexFormula1" title="10\sqrt{x}
AveGali [126]
Factor the expression
Use the properties of radicals
Simplify the roots
Calculate
And your solution will be 20 overall 5
Message me if you want me to write it all down!
8 0
3 years ago
Choose the function table that matches the given rule<br> Rule:output=input -2
kirill [66]

Answer:

option B

Step-by-step explanation:

Rule:output=input -2

LEts analyze the table

we use the input and apply the rule to get the output

LEts subtract 2 from input to get the output

Input             Rule (input -2 )            Output

-3                         -3-2                          -5

-7                         -7-2                          -9

6                          6-2                            4

Option B matches with our table

   

7 0
3 years ago
( P + 5 ) - 4 = 6<br><br>What is the value of P?
Phoenix [80]

Answer:

ight first you add 4 to 6 and that gets you 10 so you left with p+5=10 then you subtract 5 from 10 so p is gonna be  5

Step-by-step explanation:

8 0
2 years ago
Write a polynomial function of least degree with the given zero. -1+ √ 2, √ 3
kirill115 [55]

Answer:

f(x) = x^{4} + 2x³ - 4x² - 6x + 3

Step-by-step explanation:

Note that radical zeros occur in conjugate pairs, thus

- 1 + \sqrt{2} is a zero then - 1 - \sqrt{2} is also a zero

\sqrt{3} is a zero then - \sqrt{3} is also a zero

Thus the corresponding factors are

(x - (- 1 + \sqrt{2}) ), (x - (- 1 - \sqrt{2}) ), (x - \sqrt{3}), (x - (- \sqrt{3})), that is

(x + 1 - \sqrt{2}), (x + 1 + \sqrt{2}), (x - \sqrt{3}), (x + \sqrt{3})

The polynomial is then the product of the roots

f(x) = (x + 1 - \sqrt{2})(x + 1 + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3})

     = ((x + 1)² - (\sqrt{2})²)((x² - (\sqrt{3})²)

     = (x² + 2x + 1 - 2)(x² - 3)

     = (x² + 2x - 1)(x² - 3) ← distribute

     = x^{4} - 3x² + 2x³ - 6x - x² + 3

     = x^{4} + 2x³ - 4x² - 6x + 3

6 0
4 years ago
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