HELP!!! for 20 points

Kdyknuokknhnonn :) if you get this imma marry you right here right now.

Annette [7]

Answer:

its mean marry me right here right now

8 0

3 years ago

Let G be a connnectde graph with n vertices and m edges. supposed also that m = n. prove that G contains exactly one cycle

ElenaW [278]

Answer:

Contradiction

Step-by-step explanation:

Suppose that G has more than one cycle and let C be one of the cycles of G, if we remove one of the edges of C from G, then by our supposition the new graph G' would have a cycle. However, the number of edges of G' is equal to m-1=n-1 and G' has the same vertices of G, which means that n is the number of vertices of G. Therefore, the number of edges of G' is equal to the number of vertices of G' minus 1, which tells us that G' is a tree (it has no cycles), and so we get a contradiction.

7 0

3 years ago

A football is kicked with a speed of 18.0 m/s at an angle of 36.9° to the horizontal.

nata0808 [166]

Answer:

A

Step-by-step explanation:

horizontal component=18cos 36.9°≈14.39 m/s≈14.4 m/s

Vertical component=18 sin 36.9°≈10.81 m/s≈10.8 m/s

8 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:

Taylor series 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:

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

Taylor Series summation notation:

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

Therefore:

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

7 0

2 years ago

Determine if the statement is true or false The figures below are similar

artcher [175]

Answer:

True

Step-by-step explanation:

The polygons given have 5 sides, so the sum of their interior angles is (5-2) * 180 = 540. Next, there are 4 pairs of corresponding sides. Writing those sides as a, b, c, and d, for each pentagon, we can write

540 = a + b + c + d + missing angle

As is shown here, if a, b, c, and d are the same, 540 - (a+b+c+d) = missing angle is the same as well. Therefore, there are 5 corresponding angle pairs for the two polygons. Furthermore, because there are only 5 angles for each polygon, and because having all corresponding angles be equal makes polygons similar to each other, the polygons given are similar

7 0

3 years ago