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OLga [1]
3 years ago
5

Match wash hypotenuse length with the leg lengths that will create a right triangle

Mathematics
2 answers:
Ugo [173]3 years ago
5 0
Hope this can help you

Stells [14]3 years ago
3 0

Answer:

Step-by-step explanation:

Using the Pythagoras theorem, we have

(Hyp)^2=(Leg)^2+(Leg)^2

(A) The given value of hypotenuse is:

\sqrt{6} units

Now, using the Pythagoras theorem,

\sqrt{6}=\sqrt{(\sqrt{5})^2 +(1)^2}

Thus, one leg is \sqrt{5} units and the other is 1 units.

(B)The given value of hypotenuse is:

\sqrt{5} units

Now, using the Pythagoras theorem,

\sqrt{5}=\sqrt{(\sqrt{2})^2 +(\sqrt{3})^2}

Thus, one leg is \sqrt{2} units and the other is \sqrt{3} units.

(C) The given value of hypotenuse is:

\sqrt{8} units

Now, using the Pythagoras theorem,

\sqrt{8}=\sqrt{(\sqrt{5})^2 +(\sqrt{3})^2}

Thus, one leg is \sqrt{5} units and the other is \sqrt{3} units.

(D) The given value of hypotenuse is:

\sqrt{3} units

Now, using the Pythagoras theorem,

\sqrt{3}=\sqrt{(\sqrt{2})^2 +(1)^2}

Thus, one leg is \sqrt{2} units and the other is 1 units.

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Numărul cu 289 mai mic decât 812
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Answer:

The number is 523

Step-by-step explanation:

<u><em>The question is English is</em></u>

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Let

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Simplify.
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Aidan took a 1020 km trip, travelling by Bus and By Train. The bus travelled at 105 km/h, and the train travelled at 160 km/h. T
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Answer:

  523.6 km

Step-by-step explanation:

Let x represent the distance traveled by train. Then 1020-x is the distance traveled by bus. The travel times are given by the relation ...

  time = distance/speed

Then the total travel time in hours is ...

  x/160 +(1020-x)/105 = 8

  21x +32640 -32x = 26880 . . . . .  multiply by 3360

  5760 = 11x . . . . . . . . . add 11x -26880

  523.6 ≈ x . . . . . . . . . . divide by 11

Aidan traveled about 523.6 km by train.

3 0
3 years ago
Find an equation of the line that is parallel to the graph of 2x+3y=5 and contains the point (3, −1).
uranmaximum [27]

Answer:

A) Equation of line is 3X - 2Y = 11

B) Equation of line is Y = X + 1  

C) Equation of line is Y - 7 = 0

Equation of line , point (3 , -1) and slop \frac{3}{2} is 3X - 2Y = 11

Step-by-step explanation:

Given equation of line as  2x + 3y = 5

slove of this line (m1) = - \frac{2}{3}

Now another line passes through point  (3 , -1) having slop (m2)

Both lines are parallel to each other, i.e (m1) × (m2) = - 1

So slop of second line (m2 ) = \frac{-1}{m1}

                                      (m2) = \frac{-1}{(\frac{-2}{3} } )

                                      (m2) = \frac{3}{2}

Now eq of line with point (3 , -1) and slop (m2)

Y- y1 = (m2)( X - x1)

Or, Y +1 = \frac{3}{2} (X - 3)

i.e 2Y + 2= 3X - 9

∴ Equation of line is 3X - 2Y = 11     Answer

B) Given points as ( 5,6) and (3 , 4)

Slop (m) = \frac{y2 - y1}{x2 - x1} = \frac{4 - 6}{3 - 5} = 1

So the equation of line is

Y- y1 = (m)( X - x1)

Y - 6 = (1) (X - 5)

i.e Equation of line is Y = X + 1   Answer

C) Given points as ( 0, 7)  , (0 , -8)

Slop (m) with co-ordianate =  \frac{y2 - y1}{x2 - x1}

Or,                                        =  \frac{-8 - 7}{0 - 0}

                                            =0

Hence equation of line is Y - 7 = 0     Answer

7 0
3 years ago
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