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3 years ago

Two rays with a common endpoint

2 answers:

8 0

The common endpoint of two rays at which an angle is formed. Vertical Angles - Pairs of angles formed where two lines intersect. These angles are formed by rays pointing in opposite directions, and they are congruent.

Slav-nsk [51]3 years ago

6 0

This is an angle.

start at a point and the rays shoot off.

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Evaluate the expression -434 + (-1.5)2

lorasvet [3.4K]

Answer: -437

Step-by-step explanation:

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2 years ago

Change the following improper to mixed plz help quick! 1. 8/3 2. 9/4 3. 12/5

gayaneshka [121]

Answer:

Improper Fractions:

Step-by-step explanation:

1. 2 and 2/3

2. 2 and 1/4

3. 2 and 2/5

3 0

2 years ago

PLEASE HELP ME IS DUE TMR

IrinaVladis [17]

Answer:

p = 8000 + 2(6000)

20000

Step-by-step explanation:

Given that:

Profit in 1988 = 6000

Profit in 2003 = 8000 more Than double the profit made in 1998

Hence profit in 2003 (p) can be expressed as :

p = 8000 + 2(6000)

Hence profit in 2003

p = 8000 + 12000

p = 20,000

6 0

2 years ago

In parallelogram WXYZ, what is CY? 11 ft 15 ft 21 ft 23 ft

erastovalidia [21]

Answer:

15 ft

Step-by-step explanation:

In a parallelogram the diagonals bisects each other

So WC= CY

$x+4=2x-7$ solve the equation for x

Subtract x from both sides

$x-x+4=2x-x-7$

$4=x-7$

Now we add 7 on both sides

$4+7=x-7+7$

x=11

Now we need to find CY

$CY= 2x-7$

Plug in 11 for x

$CY= 2(11)-7=22-7= 15$

6 0

3 years ago

Read 2 more answers

a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?

Kruka [31]

Find the length of the radius.

$r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50} 
\\ r=5 \sqrt{2}$

Find the length of the diameter.

d = 2r = 2 × 5√2 = 10√2

Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
\\7y-42=-x-4
\\x+7y-38=0$

The distance from B(x, y) to C(3, 5) is 5√2.
$\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} 
 \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 
\\(x-3)^2+(y-5)^2=50$

Solve system of equations.
$x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
$

$x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
$

Point B could have coordinates
$\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
$

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).

8 0

3 years ago