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3 years ago

What is the range of possible sides for side x? The other two sides are 3.2 and 5.5

Mathematics

1 answer:

Alecsey [184]3 years ago

6 0

Answer:

2.3 < x < 8.7

Step-by-step explanation:

Given 2 sides of a triangle then the 3rd side x is in the range

difference of 2 sides < x < sum of 2 sides , that is

5.5 - 3.2 < x < 5.5 + 3.2

2.3 < x < 8.7

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Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1

PolarNik [594]

Answer:

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

Step-by-step explanation:

We are given that function

$f(x)=(1+3x)^{-1/3}$

We have to find the first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

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

$f(0)=(1+3x)^{\frac{-1}{3}}=1$

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

$f'(0)=-1$

$f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}$

$f''(0)=4$

$f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}$

$f'''(0)=-28$

Substitute the values we get

$(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...$

$(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...$

First term=1

Second term=-x

Third term= $2x^2$

Fourth term = $-\frac{28}{3!}x^3$

5 0

3 years ago

the width of a rectangle is 2/3 the length. the perimeter of the rectangle is 15 centimeters. what is the length of the rectangl

tester [92]

Lenght is 4 1/2 cm
P=2L +2W
15cm = 2L + 2(2/3L)
15cm = 2L + 4/3L
15cm = 2 4/3 L
15cm ÷ 2 4/3 = 2 4/3L ÷ 2 4/3
4 1/2cm = L

Check
P=2L +2W
P=2(4.5) +2(3)
15= 9 +6
15=15

4 0

3 years ago

Solve for x. Round all answers to the nearest tenth. 9 X 12

maria [59]

Step-by-step explanation:

x=54

use pyth then the other line = 15

then 15÷90 = 9÷x

then x = 54

8 0

3 years ago

Read 2 more answers

MATH Please heeeeeeeeeeeelp

Lelechka [254]

Answer:

3.002 * 10⁵

Step-by-step explanation:

3 * 10⁵ = 3 * 10³ * 10² = 3000 *10²

3 * 10⁵ + 2* 10² = 3000 * 10² + 2*10²

= (3000 + 2) * 10²

= 3002 * 10²

= 3.002 * 10³ * 10²

= 3.002 * 10⁵

6 0

3 years ago

What sum of money amounts to rupees 545 4 at 8% BA in 30 months

Mars2501 [29]

Answer:

let sum of money be x

amount=rupees5454

rate=8%

time=30 months

convert month into year

30/12=5/2 years

Step-by-step explanation:

interest=p*t*r/100

=x*5*8/100*2

=40x/200

=x/5

so to find principal use formula

amount=principal +interest

5454=x+x/5

5454=5x+x/5

5454*5=6x

27270/6=x

4545=x

therefore ,sum of money(principal)=rupees.4545

Hope this helps u!!

5 0

3 years ago