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Setler79 [48]
3 years ago
15

Factor 8x^5+4x^2−12

Mathematics
2 answers:
Effectus [21]3 years ago
4 0
D) because only 4 is the common factor in the equation.
harina [27]3 years ago
4 0

Answer:

Option D : 4(2x^{5} +x^{2} -3)

Step-by-step explanation:

Given :8x^{5} +4x^{2} -12

Solution:

Option A: 8(x^{5} +4x^{2} -12)

solving this

8x^{5} +32x^{2} -96

Since we are not getting the given expression after solving this .

Hence this wrong option

Option B: 4(2x^{5} +4x^{2} -12)

solving this

8x^{5} +16x^{2} -48

Since we are not getting the given expression after solving this .

Hence this wrong option

Option C: 4x^{2}(2x^{3} + x -3)

Solving this

8x^{5} + 4x^{3} -3x^{2}

Since we are not getting the given expression after solving this .

Hence this wrong option

Option D : 4(2x^{5} +x^{2} -3)

Solving this

8x^{5} + 4x^{2} -12

Since we are getting the given expression after solving this option

Hence Option D is correct

You might be interested in
If K is the midpoint of JL, JK = 9x -1 and KL = 2x +27, find JL
Liula [17]

Answer:

JL = 70

Step-by-step explanation:

Since K is the midpoint of JL then JK = KL and

JL = JK + KL = 9x - 1 + 2x + 27 = 11x + 26

solve for x using JK = KL

9x - 1 = 2x + 27 ( subtract 2x from both sides )

7x - 1 = 27 ( add 1 to both sides )

7x = 28 ( divide both sides by 7 )

x = 4, hence

JL = 11x + 26 = (11 × 4 ) + 26 = 44 + 26 = 70


3 0
4 years ago
Read 2 more answers
Find the length of segment KL in the trapezoid shown.
aleksklad [387]
KLMN is the trapezoid. So:( KL + MN ) / 2 = HJ( 4 x + 1 + 27 ) / 2 =  5 x + 2    / * 24 x + 28 = 10 x + 410 x - 4 x = 28 - 46 x = 24x = 24 : 6x = 4KL = 4 * 4 + 1 = 16 + 1 = 17Answer: KL = 17

3 0
3 years ago
(a) Use the Quotient Rule to differentiate the function f(x)=tan(x)-1/sec(x). f'(x)=
yKpoI14uk [10]

Answer:

(a) sin(x) + cos(x)

(b) sin(x) + cos(x)

(c) Both answers are equivalent

Step-by-step explanation:

(a) The given function is:

f(x) = \frac{tan(x)-1}{sec(x)}

According to the quotient rule:

f(x) = \frac{g(x)}{h(x)}\\f'(x)= \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}

Applying the quotient rule:

f(x) = \frac{tan(x)-1}{sec(x)}\\f'(x)=\frac{sec^2(x)*sec(x)-(tan(x)-1)*sec(x)tan(x)}{sec(x)^2}\\f'(x)=\frac{sec^3(x)-sec(x)tan^2(x)+sec(x)tan(x)}{sec(x)^2}\\f'(x)=sec(x)+\frac{tan(x)-tan^2(x)}{sec(x)}\\ \frac{tan(x)}{sec(x)}=sin(x) \\f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\

This can be simplified to:

f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\f'(x) = \frac{1}{cos(x)}+sin(x)-\frac{sin^2(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{sin^2(x)+cos^2(x)+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{cos^2(x)+sin(x)cos(x)}{cos(x)}\\ f'(x)=sin(x) +cos(x)

(b) Simplifying in terms of sin(x) and cos(x):

f(x) = \frac{tan(x)-1}{sec(x)}\\f(x)=\frac{\frac{sin(x)}{cos(x)}-1 }{\frac{1}{cos(x)} } \\f(x)=sin(x)-cos(x)\\f'(x) = cos(x)+sin(x)

(c) As proven above, both answers are equivalent.

4 0
3 years ago
How can you break up the figure into familiar shapes to determine the area?
il63 [147K]
Break it off where the triangle goes into a square/ rectangle
3 0
3 years ago
Read 2 more answers
What would W equal in w-293=379?
allochka39001 [22]

Answer: 672

Step-by-step explanation:

You would add 293+ 379, thats the answer then you can check it by doing 672-293 to see if it = 379 which it does :D

8 0
4 years ago
Read 2 more answers
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