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rusak2 [61]
3 years ago
13

Which ordered pair is a solution of the equation?

Mathematics
1 answer:
olasank [31]3 years ago
6 0

Answer:C

Step-by-step explanation:

cause I say so.

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In the right △ABC, CD is the altitude to the hypotenuse AB and m∠ABC=30°. Find AD, if BD=54 cm.
LUCKY_DIMON [66]

The ratios of sides in a 30°-60°-90° triangle are such that the hypotenuse (AB) is double the length of the shortest side (AC). In your triangle, AC is the hypotenuse of similar triangle ACD, which has AD as its shortest side.

Then

... (1/2)×(AD+54) = AC

... (1/2)×AC = AD

Together, these tell you

... AD = (1/4)(AD +54)

... 3×AD = 54

... AD = 18

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3 years ago
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Y= -1 <br> -2x-y=3<br> Solve by substitution
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I think it might be 3.
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Which plane figure generates a cylinder when it rotates about the dashed line?
MA_775_DIABLO [31]
The circle will form a sphere. (See 1st figure attached)

The rhombus (it can also be a square) will form a 2 cones attached at the bottom. (See 2nd figure attached)

The rectangle will form a cylinder. (See 3rd figure attached)

The triangle will form a cone.


Answer: The third figure.

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3 years ago
Suppose you can skin a cat using either the Cox Method or the Wong Way. In the Cox Method, there are 2 steps: Step 1 can be done
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Answer:Looking at myy profile u know im dumb

Step-by-step explanation:

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3 years ago
Graph the line and the circle (x-1)²+(y-2)²=2² and y=2x+2.
Temka [501]

Answer:

The graph of the line and the circle is in the attachment.

Step-by-step explanation:

In order to graph the circle, you have to use the circle equation formula to obtain the center and the radius.

(x-xo)² + (y-yo)² = r²

where the point (xo,yo) is the center and r is the radius.

Using the given formula, the center is:

xo=1, yo=2

(1,2)

And the radius is:

r=2

Now you have to graph the point (1,2) and draw a circle centered at that point, with a distance of 2 from the center.

To obtain the exact values of the intersection of the circle with the x-axis you have to replace y=0 in the equation and solve it for x. For the points of intersection with the y-axis you have to replace x=0 in the equation and solve it for y.

-For x=0, solving for y:

(0-1)²+(y-2)²=2²

1+(y-2)²=4

(y-2)²= 3

Applying the squareroot both sides

y-2 = ± √3

y1= -√3 +2

y2= √3 +2

P1(0, -√3 +2) and P2(0, √3 +2)

-For y=0

(x-1)²+(0-2)²=2²

(x-1)² + 4 = 4

(x-1)² = 0

Applying the squareroot both sides:

x-1 = √0

x-1=0

x=1

P3 (1,0)

In order to graph the linear equation, you have to obtain the intersection with the x-axis replacing y=0 in the equation and the intersection with the y-axis replacing x=0 in the equation. That way, you obtain two point of the line and then you have to trace the line containing those points.

-For x=0

y=2(0)+2 =2

P1 (0,2)

-For y=0

0= 2x+2

Adding -2 both sides

2x= -2

Dividing by 2 both sides

x= -1

P2(-1,0)

5 0
3 years ago
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