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Lisa [10]
3 years ago
11

HELP ASAP! Algebra II Questions!!

Mathematics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

The answer to your question is: the last option   5a² + 3b + 6a

Step-by-step explanation:

                            7a² + 3b + 6a - 2a²

 look for like terms

                           7a² - 2a²        3b         6a

Simplify like terms

                          5a² + 3b + 6a

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12a^3b^2c
Advocard [28]

Answer:

b. the degree is 3

Step-by-step explanation:

degree is the highest exponent

6 0
3 years ago
Here try to figure out this problem
Marysya12 [62]

Answer:

2

Step-by-step explanation:

2 is the slope, it's going up 2, and right 1 which is 2/1 or just 2

Hope this helps

7 0
3 years ago
Last year, Mr. Engle's total income was $52,000, while his total expenses were $53,800. Use the expression I-E/12, where I repre
olga55 [171]
Given Mr. Engle’s total income, I = $52,000 and total expenses, E = $53,800, solving for average difference between income and expenses per month:
(I-E)/12 = ($52,000 - $53,800)/12
(I-E)/12 = - $1,800/12
(I-E)/12 = -$150
This means that Mr. Engle’s expenses exceed his income by an average of $150 each month in the said year.
3 0
3 years ago
Read 2 more answers
A) Determine the functions whose graphs are parallel to the graph of the function y=0.5x+10 and intersect the graph of the funct
Elza [17]

Answer:

y - 1 = 1/2(x -5)

Step-by-step explanation:

I don't know i am right tho so be careful

7 0
2 years ago
A particle sits on a smooth surface and is acted upon by a time dependent horizontal force, giving it an
garik1379 [7]

(a) By the fundamental theorem of calculus,

<em>v(t)</em> = <em>v(0)</em> + ∫₀ᵗ <em>a(u)</em> d<em>u</em>

The particle starts at rest, so <em>v(0)</em> = 0. Computing the integral gives

<em>v(t)</em> = [2/3 <em>u</em> ³ + 2<em>u</em> ²]₀ᵗ = 2/3 <em>t</em> ³ + 2<em>t</em> ²

(b) Use the FTC again, but this time you want the distance, which means you need to integrate the <u>speed</u> of the particle, i.e. the absolute value of <em>v(t)</em>. Fortunately, for <em>t</em> ≥ 0, we have <em>v(t)</em> ≥ 0 and |<em>v(t)</em> | = <em>v(t)</em>, so speed is governed by the same function. Taking the starting point to be the origin, after 8 seconds the particle travels a distance of

∫₀⁸ <em>v(u)</em> d<em>u</em> = ∫₀⁸ (2/3 <em>u</em> ³ + 2<em>u</em> ²) d<em>u</em> = [1/6 <em>u</em> ⁴ + 2/3 <em>u</em> ³]₀⁸ = 1024

7 0
3 years ago
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