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2 years ago

Answer this 2(3^2 + 4^2) = ?

Mathematics

1 answer:

Vikentia [17]2 years ago

6 0

Answer:

Step-by-step explanation:

2(3^2+4^2)=

3^2/(3x3) = 9

4^2/(4x4) = 16

9+16=25

so.... rewritten it is

2(25)= or 2 x 25= 50

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3273 is least or greather to 3279

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3279 is greater than 3273

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2 years ago

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The sum of two numbers is 18 and the difference is 8

Andrews [41]

Answer:

x = 13 and y = 5

Step-by-step explanation:

x + y = 18

x - y = 8

Step 1: Find y

x = 18 - y

(18 - y) - y = 8

18 - y - y - 18 = 8 - 18

-2y / -2 = -10 / -2

y = 5

Step 2: Plug into x + y = 18 to get x

x + 5 - 5 = 18 - 5

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2 years ago

The ordered pair below is a solution to which equation? (2, 12) A.y=1/5x B.y=1/2x C. y = 2x D. y = 6x

Orlov [11]

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2 years ago

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Use the net to find the lateral area of the cylinder 9yd 4yd

NNADVOKAT [17]

The formula of a lateral area of a cylinder:

$LA=2\pi rH$

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3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

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2 years ago