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Viktor [21]
3 years ago
13

Is 36 km greater than less than or equal to 36,000 m

Mathematics
2 answers:
guapka [62]3 years ago
6 0
The answer is the same
Artist 52 [7]3 years ago
3 0
Every km is worth 1000 m
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The 2 in 324 represents the number 20 out of 300 and 4. With decimal in front, the 2 now represents hundredths of 1.
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Solve the system of equations using substitution. y = 2x – 10 y = 4x – 8
Dmitry_Shevchenko [17]
If this paper dosent help u solve it on Your own i can walk u through it

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Which polygon has perpendicular sides?
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c

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3 years ago
The area of a right triangle is sixty square centimetres. Its base is one centimetre less than twice its height. If the base and
NemiM [27]

Answer:

5. None of the above

Step-by-step explanation:

We have this information:

A = 60 cm2

b = 2*h - 1

Let's find out b and h:

The equation of the are of a triangle is A = (b*h)/2, if we replace it with our information, we have

60 = ((2*h - 1) * h)/2\\ 120 = 2*h^2 - h\\ 0 = 2*h^2 - h - 120

Let's find h with the quadratic formula:

\fbox {Quadratic formula:}  \frac{-b\pm\sqrt{b^2-4ac}}{2a}

h = \frac{-(-1)\pm\sqrt{(-1)^2-4*2*(-120)}}{2*2} = \frac{1\pm\sqrt{961}}{4} = \frac{1\pm\ 31}{4}

h = 8 or h = -7.5

But h represents the height of the triangle, so it has to be a positive number, that's h = 8.

If we replace this in the equation we had for b, we have that b = 2*8 - 1 = 15.

Now we can calculate the hypotenuse with the Pythagorean equation  

\fbox {Pythagorean equation:} <em>The square of the length of the hypotenuse (the side opposite the right angle) of a right triangle is equal to the sum of the squares of the two legs (the two sides that meet at a right angle). </em>

The base and the height are our legs. We will use "H" for the hypotenuse

H^2 = b^2 + h^2 = 15^2 + 8^2 = 289\\ H = \sqrt {289} = 17

H = 17

If we decrease the base and the height by 2 centimeters, we have  

b' = 15 - 2 = 13 and h' = 8 - 2 = 6

With this, let's calculate the new hypotenuse:

H'^2 = b'^2 + h'^2 = 13^2 + 6^2 = 205\\ H' = \sqrt {205} \approx 14.3

H' \approx 14.3

So, the hypotenuse decreases H - H' \approx 17 - 14.3 = 2.7

3 0
3 years ago
I need help on #10
Ede4ka [16]
D = sq.root of((x2 - x1)^2 + (y2 - y1)^2)
sq.root of 13 = sq.root of((x - (-2))^2 + (1 - 3)^2)
= sq.root of((x + 2)^2 + (-2)^2)
= sq.root of (x^2 + 4x + 4 + 4)
= sq.root of (x^2 + 4x + 8)

Now if we square both sides:
x^2 + 4x + 8 = 13
x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -5 or x = 1
6 0
3 years ago
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