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oksian1 [2.3K]
3 years ago
11

Hello, this is what I’ve got so far, please help me somebody!

Mathematics
1 answer:
topjm [15]3 years ago
7 0

Answer:

6 .... 1

3 ... 1/100

9.... 1/1000

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What is the simplified form of x+9/2x+3+x+4/x+2?
Mice21 [21]
The answer is C because multiply and divide by (x+2)(2x+3) 
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= (3x^2+22x+30) / (2x^2+7x+6) 
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Find the product.<br> (x2+6x +9)(x?) -<br> What’s the product
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Answer:

Step-by-step explanation:

(x² + 6x + 9)(x)= x²*x + 6x*x + 9*x = x³ + 6x²  + 9x

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Write the linearization of the function at the points indicated. (Enter your answer as an equation. Let x be the independent var
Natalija [7]

Answer:

\displaystyle y=5+\frac{x}{10}

\displaystyle y=10+\frac{(x-75)}{20}

Step-by-step explanation:

<u>Linearization</u>

It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.

Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is

y=f(x)=f(a)+f'(a)(x-a)

Let's find the linearization for the function

y=\sqrt{25+x}

at (0,5) and (75,10)

Computing f'(x)

\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}

At x=0:

\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}

We find f(0)

f(0)=\sqrt{25+0}=5

Thus the linearization is

\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x

\displaystyle y=5+\frac{x}{10}

Now at x=75:

\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}

We find f(75)

f(75)=\sqrt{25+75}=10

Thus the linearization is

\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)

\displaystyle y=10+\frac{(x-75)}{20}

5 0
3 years ago
Find the value of the cylinder 6 4
levacccp [35]

Answer:

9 inches ?

Step-by-step explanation:

im not 100% sure

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Find the surface area of the triangular prism??
AnnyKZ [126]

Answer:

5120 cm^2

Step-by-step explanation:

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Base perimeter = 18 + 41 + 41 = 100 cm

LA =  100 x 44 = 4400 cm^2

surface area = 360 x 2 + 4400 = 5120 cm^2

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3 years ago
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