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2 years ago

100 points! Mhanifa please help. Look at the picture attached. I will mark brainliest!

Mathematics

2 answers:

aniked [119]2 years ago

7 0

Answer:

BC = 54 units
m∠KJL = 22°

Step-by-step explanation:

BD is the perpendicular bisector of AC because AD = DC and DB⊥AC, therefore AB = BC

6x = 9x - 27
9x - 6x = 27
3x = 27
x = 9

BC = 9x - 27 = 9*9 - 27 = 81 - 27 = 54

-----------------------

JL is angle bisector of KJM, because KL = ML and KL⊥JK & ML⊥JM, therefore

m∠KJL = m∠MJL
4x + 6 = 3x + 10
4x - 3x = 10 - 6
x = 4

m∠KJL = 4*4° + 6 = 22°

PIT_PIT [208]2 years ago

4 0

Answer:

BC=54

m<KJL=22°

Step-by-step explanation:

6x=9x-27

27=3x

x=9

9(9)-27=54

BC=54

4x+6=3x+10

x=4

4(4)+6=22

m<KJL=22°

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Kenneth loaned $1,200 to his niece to buy a computer. Two years later, she paid him back the $1,200 along with an interest of $9

insens350 [35]

Answer:

The annual interest rate was 4%

Step-by-step explanation:

1. Let's review the data given to us for solving the question:

Loan to Kenneth's niece to buy a computer = US$ 1,200

Duration of the loan = 2 years

Niece paid back to Kenneth = US$ 1,296

Annual interest rate = x%

2. Let's find the annual interest rate after 2 years, using the following formula:

FV = PV * (1 + x)

PV = Loan to Kenneth's niece to buy a computer = US$ 1,200

FV = Niece paid back to Kenneth = US$ 1,296

rate (r) = x

Replacing with the real values:

1,296 = 1,200 * (1 + x)

1.296/1.200 = (1 + x) (Dividing by 1,200 at both sides)

1.08 = 1 + x

1.08 - 1 = x (Subtracting 1 at both sides)

0.08 = x

This is the interest rate for two years, for finding the annual interest rate we divide it by 2, this way:

0.08 / 2 = 0.04 * 100 = 4%

7 0

3 years ago

"let v = r 2 with the usual addition and scalar multiplication defined by k(u1, u2) = (ku1, 0). determine which of the five axio

expeople1 [14]

Let $\mathbf u\in\mathbb R^2$ , where

$\mathbf u=(u_1,u_2)$

and let $k\in\mathbb R$ be any real constant.

Given this definition of scalar multiplication, we can see right away that there is no identity element $e$ such that

$e\mathbf u=\mathbf u$

because

$e\mathbf u=e(u_1,u_2)=(eu_1,0)\neq(u_1,u_2)=\mathbf u$