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Fiesta28 [93]
3 years ago
5

HELP PLEASE solve the right triangle

Mathematics
1 answer:
aleksandrvk [35]3 years ago
7 0

Answer:

Part 1) FG=4.4\ units

Part 2) EF=3.9\ units

Part 3) m\angle G=63^o

Step-by-step explanation:

step 1

Find the measure of length side FG

In the right triangle EFG

we know that

sin(27^o)=\frac{GE}{FG} ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

sin(27^o)=\frac{2}{FG}

FG=\frac{2}{sin(27^o)}=4.4\ units

step 2

Find the measure of length side EF

In the right triangle EFG

we know that

cos(27^o)=\frac{EF}{FG} ----> by CAH (adjacent side divided by the hypotenuse)

substitute the given values

cos(27^o)=\frac{EF}{4.4}

EF=cos(27^o)(4.4)=3.9\ units

step 3

Find the measure of angle G

we know that

m\angle G+27^o=90^o ---> by complementary angles in a right triangle

m\angle G=90^o-27^o=63^o

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Nina [5.8K]

Answer:

a) 3 b) 5 c) 7 d) 9

Step-by-step explanation:

For this, you want to replace x in the equation y=x+5 with each of the values listed.

For the first one, -2, the equation becomes y=-2+5, which is solved to y=3.

For the second one, 0, the equation becomes y=0+5, which is solved to y=5.

For the third one, 2, the equation becomes y=2+5, which is solved to y=7.

For the last one, 4, the equation becomes y=4+5, which is solved to y=9.

**This content involves solving algebraic equations with a known variable, which you may wish to revise. I'm always happy to help!

5 0
3 years ago
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Step-by-step explanation:

3 0
3 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
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tatyana61 [14]

Answer: D, an equilateral triangle has three congruent sides.

4 0
3 years ago
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Luda [366]

Answer:

B, C, E are true

Step-by-step explanation:

A = l*w

The original rectangle is 6 by 4  with A = 24

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The second enlargement is 18*3 by 12*3

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A.The area of the first enlargement is 72 square inches.

False it is 216

B.The area of the second enlargement is 1,944 square inches.

True

C.The area of the second enlargement is (3 squared) squared times the original area.

216/24 =9  True

D.The area of the second enlargement is 3 times the area of the first enlargement.

1944/24 = 9  False

E.The ratio of the area of the first enlargement to the area of the original equals the square of the scale factor.

The scale factor is 3 3^2 =9 216/24 =9

True

4 0
3 years ago
Read 2 more answers
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