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Bas_tet [7]
2 years ago
10

Write an equation of the line containing the given point and parallel to the given line.

Mathematics
1 answer:
nadya68 [22]2 years ago
4 0

Answer:

Solution given:

9x-2y=7

making it in a form of y=mx+c

9x-7=2y

y=9/2x -7/2

Comparing above equation with y=mx+c

we get

m=9/2

Since it is passes parallel to (5,-3)

slope will be same

now

Equation of line

y-y1=m(x-x1)

y+3=9/2(x-5)

y=9/2x-5/2-3

<u>y=9/2 x-11/2 is a required equation.</u>

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I need the explanation and the answer of 14/40 in simplest form
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Answer:

\frac{7}{20}

Step-by-step explanation:

To simplify a fraction, you need to find the GCF (Greatest Common Factor) of the numerator (top number) and denominator (bottom number).

Factors of 14:

1, 2, 7, 14

Factors of 40:

1, 2, 4, 5, 8, 10, 20, 40

As you can see, 2 is the GCF of 14 and 40. So, we now divide both numbers by 2.

14 ÷ 2 = 7

40 ÷ 2 = 20

Therefore, \frac{14}{40} in simplest form is \frac{7}{20}

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A tv has height:width in ratio 9:16. In a scale drawing the height is 4.5cm. What would the width be?
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Find the third order maclaurin polynomial. Use it to estimate the value of sqrt1.3
vodka [1.7K]

\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3} is the maclaurin polynomial and estimate value of \sqrt{1.3} is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

<h3>Find the third order maclaurin polynomial:</h3>

Given the polynomial,

f(x)=\sqrt{1+3x}=(1+3x)^{\frac{1}{2} }

The formula to find the maclaurin polynomial,

f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}

Next we have to find f'(x), f''(x) and f'''(x),

  • f'(x) = \frac{3}{2}(1+3x)^{-\frac{1}{2} }
  • f''(x) =-\frac{9}{4}(1+3x)^{-\frac{3}{2} }
  • f'''(x) = \frac{81}{8}(1+3x)^{-\frac{5}{2} }

By putting x = 0 , we get,

  • f(0)=(1+3(0))^{\frac{1}{2} }=1
  • f'(0) = \frac{3}{2}(1+3(0))^{-\frac{1}{2} }=\frac{3}{2}
  • f''(0) =-\frac{9}{4}(1+3(0))^{-\frac{3}{2} }=-\frac{9}{4}
  • f'''(0) = \frac{81}{8}(1+3(0))^{-\frac{5}{2} }=\frac{81}{8}

Therefore the maclaurin polynomial by using the formula will be,

\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}

\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}

To find the value of \sqrt{1.3}  we can use the maclaurin polynomial,

\sqrt{1.3} is  \sqrt{1+3x} with x = 1/10,

\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}

\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14

Hence \sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3} is the maclaurin polynomial and estimate value of \sqrt{1.3} is 1.14.

Learn more about maclaurin polynomial here:

brainly.com/question/24188694

#SPJ1

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