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SSSSS [86.1K]
3 years ago
12

Find the remainder when f(x)=2x3-5x2+8x+4 is divided by 2x-1 and use it to determine if 2x-1 is a factor

Mathematics
1 answer:
ollegr [7]3 years ago
5 0

The remainder of f(x) = 7

Step-by-step explanation:

f(x) = 2x^{3}  - 5x^{2} +8x + 4

Here 2x - 1 is a factor.

We will apply Factor theorem ( states that if (x-a) is a factor of f(x) then f(a) is the remainder)

Putting x = 1/2 in f(x) we will get the remainder.

f(1/2) = 2(1/2)^3 -5(1/2)^2 +8(1/2) +4

=7

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Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab
Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

<u>Left part:</u>

\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

<u>Right part:</u>

\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

Since simplified left and right parts are the same, then the equality is true.

3 0
3 years ago
??????????????????????
gtnhenbr [62]

Answer:

110 degrees

Step-by-step explanation:

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3 years ago
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2 years ago
If a=(-1, -3) and b=(11, -8) what is the length of ab?
professor190 [17]

Answer: D.  The length is 13 units.

Step-by-step explanation:

(-1, -3 )    Find the difference between the x values and y values.

 (11, -8)  

-3-(-8) = 5

-1-11 =   -12    now find the  absolute values.

The absolute value of 5 is 5 .  The absolute value of  -12 is 12.

Now use the Pythagorean Theorem formula a^2 + b^2 =c^2  where c square is the length between AB.

5^2 + 13^2 = C^2

25 + 144 = c^2  

169=c^2

c= 13

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3 years ago
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This is what u got: 2•8=16/2=8
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2 years ago
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