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Crank
2 years ago
12

C с 5 km 12 km What is the length of the hypotenuse? kilometers

Mathematics
2 answers:
inysia [295]2 years ago
8 0

Answer:

13km

Step-by-step explanation:

Sedaia [141]2 years ago
4 0

Answer:

c=13km

Step-by-step explanation:

13 (by pythagorean triples)

or you could use the pythagorean theorem to find c:

5^2+12^2=c^2

25+144=c^2

c^2=169

c=13

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4 times a number is 16​
andrey2020 [161]

Answer:

4 times 4 equals 16

Step-by-step explanation:

4 times 1= 4

4 times 2= 8

4 times 3= 12

4 times 4= 16

you add 4 after each result.

4 0
3 years ago
Given:
diamong [38]

Answer:

10-5\sqrt{2}

Step-by-step explanation:

As per the attached figure, right angled \triangle MDL has an inscribed circle whose center is I.

We have joined the incenter I to the vertices of the \triangle MDL.

Sides MD and DL are equal because we are given that \angle M = \angle L = 45 ^\circ.

Formula for <em>area</em> of a \triangle = \dfrac{1}{2} \times base \times height

As per the figure attached, we are given that side <em>a = 10.</em>

Using pythagoras theorem, we can easily calculate that side ML = 10\sqrt{2}

Points P,Q and R are at 90 ^\circ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of  \triangleMIL, \triangleMID and \triangleDIL.

Also,

\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL

\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2

So, radius of circle = 10-5\sqrt2

8 0
2 years ago
Nancy walks 5 blocks to the right , 3 blocks down , 7 blocks to the left and 1 block up each square size length is 1 block. What
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Answer:

b

Step-by-step explanation:

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2 years ago
How do you solve -1=-x+2
Arturiano [62]

-1 = x + 2

-2       -2

-3 = x

x = -3

8 0
2 years ago
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