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Studentka2010 [4]
2 years ago
5

Facorise this exprection fully as possible 2x^2+6x

Mathematics
2 answers:
Gre4nikov [31]2 years ago
8 0

Answer:

2x(x+3)

Step-by-step explanation:

We'll start by doing simple things.

We see that there is a common factor of 2 so we will first take that out:

2(x^2+3x)

Now we see that we can also take out an x:

2x(x+3)

Now we can't take out anything else. Done!

Butoxors [25]2 years ago
7 0

Answer:

2x(x+3)

Step-by-step explanation:

To factor, you need ti find what's the greatest common factor in your expression.

in this case, the expression is divisible by 2x, since both parts of the equation are divisible by 2x

when you take out 2x, you are left with x+3. there's nothing else to factor, so this is the simplest form:

2x (x + 3)

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Plsss help!! In triangle ABC, AC=13, BC=84, and AB=85. Find the measure of angle C
Ivahew [28]

Answer:

90°

Those three sides make a right triangle,

√(84²+13²) = 85

So angle C is 90°

Answered by GAUTHMATH

5 0
3 years ago
Brainlest- Please please help me thank you so much in advance
ExtremeBDS [4]

Answer:

x= 94

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
If n is even, which of the following cannot be odd?
lys-0071 [83]
B) 3n
well, first of all, you can try it out with numbers like 4,6,28...
3(4)
=12
3(6)
=18
3(28)
=84

also, you can disprove the other theories
a) 0+3=3 (odd)

b) 4^2-1=15 (odd)



4 0
3 years ago
Use integration by parts to derive the following formula from the table of integrals.
emmasim [6.3K]

Answer:

I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)

Step-by-step explanation:

for

I= ∫x^n . e^ax dx

then using integration by parts we can define u and dv such that

I= ∫(x^n) . (e^ax dx) = ∫u . dv

where

u= x^n → du = n*x^(n-1) dx

dv= e^ax  dx→ v = ∫e^ax dx = (e^ax) /a ( for a≠0 .when a=0 , v=∫1 dx= x)

then we know that

I= ∫u . dv = u*v - ∫v . du + C

( since d(u*v) = u*dv + v*du → u*dv = d(u*v) - v*du → ∫u*dv = ∫(d(u*v) - v*du) =

(u*v) - ∫v*du + C )

therefore

I= ∫u . dv = u*v - ∫v . du + C = (x^n)*(e^ax) /a - ∫ (e^ax) /a * n*x^(n-1) dx +C = = (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C

I= (x^n)*(e^ax) /a - n/a ∫ (e^ax) *x^(n-1) dx +C (for a≠0)

5 0
3 years ago
No links, provide explanation.
a_sh-v [17]

Answer:

C: 18

Step-by-step explanation:

6 0
3 years ago
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