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2 years ago

PLEASE HELP!!! this test will bring my grade up!! (question in picture)

Mathematics

1 answer:

kipiarov [429]2 years ago

5 0

Answer:

Step-by-step explanation:

I think it's SAS

It is because O is the common angle and the two sides are similar.

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Solve by factoring. x2-49 = 0 X=

Archy [21]

24 1/2 or 24.5 is the answer

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2 years ago

Read 2 more answers

Express sin O as a fraction in simplest terms.

ollegr [7]

To express sin O as a fraction in simplest term is 5/6

<h3>How to determine the expression</h3>

Using the formula:

Sin O = opposite/ hypotenuse

Where opposite = The side directly opposite the angle

hypotenuse = The longest side of the triangle

Sin O = 5 / 6

Therefore, to express sin O as a fraction in simplest term is 5/6

Learn more about the trigonometry here:

brainly.com/question/7331447

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1 year ago

B B' 12 7.5 8 5 А C A' C' 5 7.5 A 1

scoundrel [369]

Answer:

For me to answer this question I need to understand it, I really don't understand your question.

what is the question?

is there any diagram you left out in this "question" ?.

All I'm saying is that the question isn't clear.

thanks.

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1 year ago

I need help with this problem ASAP

Kamila [148]

Answer:

Step-by-step explanation:

4 0

2 years ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago