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3 years ago

What is the 10th term of 6,10,14,18?

Mathematics

2 answers:

Nikitich [7]3 years ago

6 0

Good evening

y = 4x + 2
y = 4×10 + 2
y = 42

the 10th term is 42

I hope I helped you

Tasya [4]3 years ago

4 0

Y = 4x + 2
y = 4×10 + 2
y = 42

the 10th term is 42

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If the length of the hypotenuse of a right triangle is 39 millimeters and the length of one leg is 15 millimeters what is the le

Genrish500 [490]

Answer:

36 millimeters

Step-by-step explanation:

From Pythagoras theorem, the square of the hypotenuse is equal to the sum of the square of the two other legs

In mathematical terms;

a^2 = b^2 + c^2

Let a represent the hypotenuse = 39 mm and the length of one leg, say c, is 15 mm

Slotting in the values of a and b

39^2 = b^2 + 15^2

1521 = b^2 + 225

collect like terms

1521 - 225 = b^2

1296 = b^2

Take the square root of both sides

36 = b

Therefore b = 36 mm

6 0

3 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$

And,

$f'(A)=\frac{dB}{dA}=\sqrt{3}$ from part b

Therefore, the linear approximation at $A=\frac{\pi}{4}$ is,

$f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}$

Use part (c), when $A=46°$ , B is approximately,

$B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°$

8 0

3 years ago

Hi i just need help checking if my answer is correct I have been working on a implicit diffrention problem and I need some help.

SSSSS [86.1K]

Answer:

$\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}$

Step-by-step explanation:

Normally, when I do this, I differentiate each term first with respect to x then with respect to y. In this solution, I differentiated the entire expression with respect to x, then with respect to y. That makes separating the dx and dy coefficients much easier, so the solution almost falls into your lap.

$-x^4 y + 2 x y^3 + x^3 y^3=0 \qquad\text{given}\\\\(-4x^3y+2y^3+3x^2y^3)dx+(-x^4+6xy^2+3x^3y^2)dy=0\\\\(-4x^3y+2y^3+3x^2y^3)dx=(x^4-6xy^2-3x^3y^2)dy\\\\\boxed{\dfrac{dy}{dx}=\dfrac{-4x^3y+2y^3+3x^2y^3}{x^4-6xy^2-3x^3y^2}}$