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3 years ago

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The figure contains a triangle. One side is 3.8 feet. A second side is 4. 3 feet. The angle between the given sides is 53 degree

s. What is the area?

1 answer:

Romashka [77]3 years ago

4 0

Area= (1/2) ab sin c
Area= (1/2) (3.8)(4.3) sin53
Area= 6.524852 answer found from ti-84

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Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

$\frac{d}{dx}\left(3x+x^3\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

solving

$\frac{d}{dx}\left(3x\right)$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'$

$=3\frac{d}{dx}\left(x\right)$

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now solving

$\frac{d}{dx}\left(x^3\right)$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}$

$=3x^{3-1}$

$=3x^2$

Thus, the expression becomes

$\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

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Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 = 3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

Thus, the value of x will be:

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5 0

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9514 1404 393

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14 units

Step-by-step explanation:

The angle bisector divides the sides proportionally.

BD/BA = CD/CA

(x+4)/8 = (2x+1)/12

3(x +4) = 2(2x +1) . . . . . multiply by 24 (the least common denominator)

3x +12 = 4x +2 . . . . . . . eliminate parentheses using the distributive property

12 = x +2 . . . . . . . subtract 3x

10 = x . . . . . . . . . subtract 2

Then the length of BD is ...

BD = x +4

BD = 10 +4 . . . . substitute the value of x

BD = 14

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