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atroni [7]
3 years ago
14

Add. (3x2 - 2x) + (4x-3)

Mathematics
2 answers:
polet [3.4K]3 years ago
7 0
I think it would be 3x2 +2x -3
ra1l [238]3 years ago
6 0

Answer:

3x^2 +2x−3

Step-by-step explanation:

3x^2 −2x+4x−3

Combine −2x and 4x to get 2x.

3x^2 +2x−3

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1. Find the compound interest and maturity value if P = P43,000, with a rate of 5% is compounded semi-annually for 6 years.​
Leno4ka [110]

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7 0
2 years ago
A=2,b=1 and c=4.What is 3ac-a+2b
kati45 [8]

Step-by-step explanation:

a = 2

b = 1

c = 4

<h2>Question:</h2>

3ac - a + 2b

= Solution ,

= 3 × 2 × 4 - 2 + 2 × 1

= 24 - 2 + 2

= 24 + 2 - 2

= 26 - 2

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6 0
2 years ago
All zeros of polynomial function 3x^4 +14x^2 -5
Marizza181 [45]

Answer:

Correct answer:  x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

Step-by-step explanation:

Given:

3 x⁴ + 14 x² - 5 = 0   biquadratic equation

this equation is solved by a shift  x² = t and get:

3 t² + 14 t - 5 = 0

t₁₂ = (-14 ± √14² - 4 · 3 · 5) / 2 · 3 = (-14 ± √196 + 60) / 6

t₁₂ = (-14 ± √256) / 6 = (-14 ± 16) / 6

t₁ = -5   or  t₂ = 1 / 3

the solution t₁ = -5 is not accepted because it cannot be x² = -5

we accepted  t₂ = 1 / 3

x² =  1 / 3  ⇒  

x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

God is with you!!!

5 0
3 years ago
After jumping rope for 15 minutes, Jan took a break. If Jan took a break at 3:05 p.M., at what time did she start jumping rope?
Kitty [74]

Answer: 2 : 50 pm

Step-by-step explanation:

If Jan took a break at 3:05 pm and she had been jumping rope for 15 minutes before that, to find the time she started jumping rope, subtract 15 from the current time:

= 3:05 - 15

= 2:50 pm

<em>Jan started jumping rope at 2 : 50 pm </em>

3 0
3 years ago
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