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AleksandrR [38]
2 years ago
5

Help mee plz... i ' m in troubleans 2,3&4

Mathematics
1 answer:
motikmotik2 years ago
5 0

Step-by-step explanation:

2) a= -3/8 and b= -5/3

a×b= b×a

<u>-3</u><u> </u>× <u> </u><u>-5</u><u> </u> = <u> </u><u>-5</u><u> </u> × <u> </u><u>-3</u><u> </u>

8. 3. 3. 8

<u> </u><u>1</u><u>5</u><u> </u> = <u> </u><u>1</u><u>5</u><u> </u>

24. 24

3)a=8/11 and b= -6/11

a×b=b×a

<u> </u><u>8</u><u> </u> × <u> </u><u>-6</u><u> </u> = <u>-6</u><u> </u> × <u> </u><u>8</u><u> </u>

11. 11. 11. 11

<u> </u><u>-48</u><u> </u> = <u> </u><u>-48</u><u> </u>

121. 121

4) a= -9/15 and b= -7/2

a×b=b×a

<u> </u><u>-9</u><u> </u> × <u> </u><u>-7</u><u> </u> = <u> </u><u>-7</u><u> </u> × <u> </u><u>-9</u><u> </u>

15. 2. 2. 15

<u> </u><u>6</u><u>3</u><u> </u> = <u> </u><u>6</u><u>3</u><u> </u> , let's divide both by 3

30. 30

<u> </u><u>2</u><u>1</u><u> </u>= <u> </u><u>2</u><u>1</u><u> </u>

10. 10

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Please help me! 15 points
arsen [322]

Answer:

a. The length of FA is 1.

b. The length of the radius (AC) is 2.

c. The circumference of the circle A is 4π units.

d. The measure of the minor arc BC is 107°.

e. The length of the minor arc BC is (107/90) π.

f. The m<BFC and m<EFD is 125.5°.

g. The m<BFE is 180°.


Step-by-step explanation:

Radius: R=AB=2

FC=1

Arc CD = 180°

Minor arc BD = 73°

Minor arc EC = 36°


a. What is the length of FA?

FA=AC-FC

AC is a radius, then AC=AB→AC=2

Replacing in the known values in the equation above:

FA=2-1

FA=1


b. What is the length of the radius (AC)?

The radius AC must be equal to the radius AB, then:

AC=AB→AC=2


c. What is the circumference of the circle A?

Circunference of circle A: L=?

L=2 π R

L=2 π (2)

L=4π


d. What is the measure of the minor arc BC?

Minor arc BC = arc CD - Minor arc BD

Minor arc BC = 180°-73°

Minor arc BC = 107°


e. What is the length of the minor arc BC?

Length of minor arc BC: l=?

l=(Minor arc BC / 360°) L

l=(107°/360°) 4π

l=(4*107/360) π

l=(107/90) π


f.  What is the m<BFC and m<EFD

<BFC and <EFD are interior angles, then:

m<BFC = m<EFD = ( Minor arc BC + Minor arc DE) / 2

Minor arc DE = arc CD - Minor arc EC

Minor arc DE = 180°-36°

Minor arc DE = 144°

m<BFC = m<EFD = ( 107° + 144° ) / 2

m<BFC = m<EFD = ( 251° ) / 2

m<BFC = m<EFD = 125.5°


g. What is the m<BFE?

<BFE is a straight angle, then m<BFE=180°





8 0
3 years ago
PLEASE HELP!
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Answer:

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Complete Question:

Find the directional derivative of g(x,y) = x^2y^5at the point (1, 3) in the direction toward the point (3, 1)

Answer:

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Step-by-step explanation:

Get g'_x and g'_y at the point (1, 3)

g(x,y) = x^2y^5

g'_x = 2xy^5\\g'_x|(1,3)= 2*1*3^5\\g'_x|(1,3) = 486

g'_y = 5x^2y^4\\g'_y|(1,3)= 5*1^2* 3^4\\g'_y|(1,3)= 405

Let P =  (1, 3) and Q = (3, 1)

Find the unit vector of PQ,

u = \frac{\bar{PQ}}{|\bar{PQ}|} \\\bar{PQ} = (3-1, 1-3) = (2, -2)\\{|\bar{PQ}| = \sqrt{2^2 + (-2)^2}\\

|\bar{PQ}| = \sqrt{8}

The unit vector is therefore:

u = \frac{(2, -2)}{\sqrt{8} } \\u_1 = \frac{2}{\sqrt{8} } \\u_2 = \frac{-2}{\sqrt{8} }

The directional derivative of g is given by the equation:

D_ug(1,3)  = g'_x(1,3)u_1 + g'_y(1,3)u_2\\D_ug(1,3)  = (486*\frac{2}{\sqrt{8} } ) +  (405*\frac{-2}{\sqrt{8} } )\\D_ug(1,3)  = (\frac{972}{\sqrt{8} } ) +  (\frac{-810}{\sqrt{8} } )\\D_ug(1,3)  = \frac{162}{\sqrt{8} }

8 0
3 years ago
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