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Mekhanik [1.2K]

3 years ago

12

X^4+4x^3+6x^2+4x+1

1 answer:

bezimeni [28]3 years ago

8 0

Factor the following:x^4 + 4 x^3 + 6 x^2 + 4 x + 1
The coefficients match the 5^th row of Pascal's triangle, so x^4 + 4 x^3 + 6 x^2 + 4 x + 1 = (x + 1)^4:Answer: (x + 1)^4

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Solve the inequality. −3y<−14

Artyom0805 [142]

Answer: y>14/3

Step-by-step explanation:

3 0

3 years ago

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Bezzdna [24]

Answer:

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7 0

3 years ago

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mr Goodwill [35]

Answer:

see explanation

Step-by-step explanation:

Using the rule of exponents

$a^{m}$ = $\frac{1}{a^{-m} }$ , then

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To divide leave first, change ÷ to × , turn second upside down

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6 0

3 years ago

The location of point V is (-3,3). The location of point X is (9,13). Determine the location of point W which is 3/4 of the way

Inga [223]

ANSWER

$W(\frac{9}{7},\frac{51}{7} )$

EXPLANATION

We want to find the coordinates of the point W(x,y) which divides V(-3,3) and X(9,13) in the ratio m:n=3:4.

The x-coordinate of this point is given by:

$x= \frac{mx_2+nx_1}{m + n}$

$x= \frac{3(9)+4( - 3)}{4 + 3}$

$x= \frac{21 - 12}{4 + 3}$

$x= \frac{9}{7}$

The y-coordinates is given by;

$y= \frac{my_2+ny_1}{m + n}$

$y= \frac{3(13)+4( 3)}{4 + 3}$

$y= \frac{39+12}{4 + 3}$

$y= \frac{51}{7}$

Hence

$W(\frac{9}{7},\frac{51}{7} )$

3 0

3 years ago

. Evaluate tan(α + β) = sin(????+????) / cos(????+????) to show tan(???? + ????) = tan(????)+tan(????????) / 1−tan(????) tan(???

grandymaker [24]

Answer with Step-by-step explanation:

We are given that

$tan(\alpha+\beta)$

$\frac{sin(\alpha+\beta}{cos(\alpha+\beta)}$

By using formula $tan x=\frac{sin x}{cos x}$

$\frac{sin\alpha cos\beta+sin\beta cos\alpha}{cos\alpha cos\beta-sin\alpha sin\beta}$

By using property: $sin(x+y)=sin x cosy+cos x sin y$

$cos(x+y)=cos x cosy-sin x siny$

Divide numerator and denominator by $cos\alpha cos\beta$

Then, we get

$\frac{\frac{sin\alpha}{cos\alpha}+\frac{sin\beta}{cos\beta}}{1-\frac{sin\alpha sin\beta}{cos\alpha cos\beta}}$

$tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}$

Hence, proved

Substitute $\alpha=\beta$

Then we get

$tan 2\alpha=\frac{tan\alpha+tan\alpha}{1-tan^2\alpha}$

$tan(2\alpha)=\frac{2tan\alpha}{1-tan^2\alpha}$

Hence, proved.

3 0

4 years ago