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lord [1]
3 years ago
9

How do I calculate the Pythagorean theorem?

Mathematics
1 answer:
kogti [31]3 years ago
7 0
A^2 + b^2 = c^2
A*2 b*2
When you get the final answer of a and b after squaring you then add a and b and you get your answer C^2
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In his science class, Boris conducted an experiment in which he used a glass prism to make a rainbow. He now wants to find the s
Norma-Jean [14]

Answer:

<u>Use the formula for area of the triangle:</u>

  • A = 1/2bh, where b- base, h- height or altitude
<h3>Part A</h3>

<u>The triangle i has:</u>

  • b = 4 cm, h = 1.5 cm

<u>The area is:</u>

  • A = 1/2*4*1.5 = 3 cm²
<h3>Part B</h3>

Triangles have same area as they are congruent.

<u>Total area is:</u>

  • 3*2 = 6 cm²
4 0
3 years ago
Find the distance between the points (4, 6) and (2, -6). Round to the nearest tenth.
Deffense [45]

Answer:

The correct answer is B.12.2

Step-by-step explanation:

5 0
3 years ago
Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere
Ket [755]

Answer:

e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...

\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...

This expansion is valid only if \text{f}\:^{(n)}(a) exists and is finite for all n \in \mathbb{N}, and for values of x for which the infinite series converges.

\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1

\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...

\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If  $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}

\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4

\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4

\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4

Substituting the values in the series expansion gives:

e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...

Factoring out e⁴:

e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]

<u>Taylor Series summation notation</u>:

\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n

Therefore:

\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n

7 0
1 year ago
Show workings please
Alenkinab [10]

Answer: x = 21°

Step-by-step explanation:

37 + 4x + (3x - 4) = 180°   Original Equation

37 + 4x + 3x - 4 = 180°   Commutative Property of Addition

7x + 33 = 180°   Combine like terms

7x = 148°  Subtract 33 from both sides

x = 21°   Divide both sides by 7

8 0
2 years ago
What is the unit price of 4 bags for $3.24
tiny-mole [99]
Unit price is $0.81 per bag
8 0
3 years ago
Read 2 more answers
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