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3 years ago

12

Which point represents 4StartRoot 2 EndRoot (cosine (StartFraction 5 pi Over 4 EndFraction) + I sine (StartFraction 5 pi over 4

Endfraction) ) ?
It's Point B

1 answer:

Alecsey [184]3 years ago

3 0

Answer: Point b

Step-by-step explanation:

They said so, thank you.

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What is the answer and the steps to get them.

Georgia [21]

Answer:

on #1 x=2 and y=2

Step-by-step explanation:

a squared + b squared = c squared

2 squared + 2 squared = 4 squared

4 + 4 = 8

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3 years ago

Marizza181 [45]

Answer:

D................ .......

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3 years ago

The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

stiks02 [169]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

$A=b^{2}$

where b is the length side of the square

we have

$A=49\ units^2$

substitute

$49=b^{2}$

$b=7\ units$

therefore

$AB=BE=ED=AD=7\ units$

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

$A=(AC)(AG)$

substitute the given values

$A=(9)(10)=90\ units^2$

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

$A=(DE)(DG)$

we have

$DE=7\ units$

$DG=AG-AD=9-7=2\ units$

substitute

$A=(7)(2)=14\ units^2$

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

$A=(EF)(CF)$

we have

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

substitute

$A=(3)(7)=21\ units^2$

step 5

sum the shaded areas

$14+21=35\ units^2$

step 6

Divide the area of of the shaded region by the area of ACIG

$\frac{35}{90}$

Simplify

Divide by 5 both numerator and denominator

$\frac{7}{18}$

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

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2 years ago

The sum of seven times a number, and 9, is equal to six times the number. What's the algebraic expression?

S_A_V [24]

7n+9=6n
9=-n
n=-9
..............

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2 years ago

JL is a diameter of circle k. If tangents to circle I are constructed through points L and K what relationships would exist betw

luda_lava [24]

i. Let t be the line tangent at point J. We know that a tangent line at a point on a circle, is perpendicular to the diameter comprising that certain point. So t is perpendicular to JL
let l be the tangent line through L. Then l is perpendicular to JL ii. So t and l are 2 different lines, both perpendicular to line JL.
2 lines perpendicular to a third line, are parallel to each other, so the tangents t and l are parallel to each other.
Remark. Draw a picture to check the

3 0

3 years ago