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Ivanshal [37]
3 years ago
6

These two triangles are congruent by the hypotenuse leg theorem.

Mathematics
1 answer:
Blababa [14]3 years ago
5 0
X-y=x+2
2x-y=4x+2y (same as we would write 2x+3y=0)

-y=2+x-x
y=-2

so 2x+3y=0 is
2x+3×(-2)=0
2x=6
x=6/2=3

x=3

answer: x=3
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Another one that is even more confusing because it is a fraction.
kogti [31]

Answer:

2/16

Step-by-step explanation:

i just doubled he equation

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3 years ago
The temperature changes from 9°F to -5°F. Which of the following expressions model the distance on thermometer between 9°F and -
scoray [572]

Answer:

d = |- 5- 9|

Step-by-step explanation:

Given

Initial = 9^\circ F

Final = -5^\circ F

Required

Determine the difference

Represent this with d.

d is calculated using

d = |Final - Initial|

The equation becomes

d = |- 5- 9|

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3 years ago
It says write in scientific notation but isn’t it already in scientific notation ???
Tems11 [23]
Oh..it sure yes hm..that’s odd
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A line includes the points (13,15) and (6,15). What is its equation in slope-intercept form?​
Andrew [12]

Answer:

y=15

Step-by-step explanation:

y=mx+b

slope(y2-y1)/(x2-x1)

(15-15)/(6-13)

0/-7=0

y-intercept 15=0(15)+b

b=15

y=15

6 0
3 years ago
The position vector of p (relative to the origin) is op = (x, y). if the magnitude of op is 5 units, find the set of all possibl
Firdavs [7]

Answer:

S = {(0, 5), (0, - 5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5,0), (-5, 0) }

Step-by-step explanation:

Remember that the distance between two points (a, b) and (c, d) is given by:

distance = \sqrt{(a - c)^2 + (b - d)^2}

So, here we have that the distance between the point (x, y) and the origin, (0, 0) must have a magnitude of 5 units, then we want to solve:

5 = \sqrt{(x - 0)^2 + (y - 0)^2} \\5 = \sqrt{x^2 + y^2}

If we square both sides, we get

5^2 = (\sqrt{x^2 + y^2} )^2\\25 = x^2 + y^2

Now we want to find all the points (x, y) that meet this condition, suc that:

x ∈ Z

y ∈ Z

So both x and y must be integers.

So here we can just play with different values of x and y.

For example, if we define:

x = 0 we get:

25 = 0^2 + y^2

25 = y^2

√25 = y

Then we can have y = 5 or  y = -5

from this we got two points:

(0, 5) and (0, - 5)

if x = 1 we have:

25 = 1^2 + y^2

25 - 1 = y^2

24 = y^2

There is no integer such that its square is equal to 24, so we can stop here.

if x = 2 or - 2, we have:

25 = 2^2 + y^2

25 = 4 + y^2

25 -4 = 21 = y^2

Again, there is no integer such that its square is equal to 21, so we can stop here.

if x = +3 or -3, we have:

25 = 3^2 + y^2

25 = 9 + y^2

25 - 9 = 16 = y^2

√16 = y

then we can have y = 4 or y = -4

from this we got four points:

(3, 4)

(-3, 4)

(3, -4)

(-3, -4)

And for symmetry, if x = 4 or -4 we have the points:

(-4, 3)

(4, 3)

(-4, -3)

(4, -3)

finally, again for symmetry, if we take x = 5 or x = -5 we have the points:

(5,0)

(-5, 0)

Concluding, the set of all possible values (x, y) is:

S = {(0, 5), (0, - 5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5,0), (-5, 0) }

7 0
3 years ago
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