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max2010maxim [7]
3 years ago
8

Find the unknown length in the right triangle 24mm 32mm

Mathematics
1 answer:
valkas [14]3 years ago
5 0
Are these two lengths legs or a leg and hypotenuse?
If both are legs:
24 squared + 32 squared = unknown leg squared

When the answer is found don't forget to find the square root

If one is a leg and another is Hypotenuse:

32 squared - 24 squared= unknown leg squared

Just plug these numbers in your calculator
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IRISSAK [1]

8 kilometers is the answer

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5 0
3 years ago
Can someone please teach me long division, and please provide examples.
Dmitrij [34]

Answer:

Step-by-step explanation:

Divide the tens column dividend by the divisor.

Multiply the divisor by the quotient in the tens place column.

Subtract the product from the divisor.

Bring down the dividend in the ones column and repeat.

okay so look at the image down below

8 0
3 years ago
Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos
Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$ P_{1} = (- 1, 4) $

$ P_{2} = (6, 2) $    and

$ P_{3} = (4, - 5) $

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

                                 $ \sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+}   \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}} $

Length between $ P_1 $ and $ P_{2} $ , (Side 1) :

$ (x_1, y_1) = (- 1, 4) $     and

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2 $

$ = \sqrt{7^2 + 2^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53}} $

Length of Side 1 = $  \sqrt{\textbf{53}} $ units.

Distance between $ P_1 $ and P_2 , (Side 2):

$ (x_1, y_1) = (-1, 4) $

$ (x_2, y_2) = (4, - 5) $

Distance = $ \sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2 $

$ = \sqrt{ 25 + 81 } $

$ = \sqrt{\textbf{106}} $

Length of Side 2 = $ \sqrt{\textbf{106}} $ units.

Distance between $ P_2 $ and $ P_3 $ , Side 3 :

$ (x_1, y_1) = (4, 5) $

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53} $

Length of Side 3  = $ \sqrt{\textbf{53}} $ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$ \bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2 $

i.e, 53 + 53  = 106

Hence, the triangle is a right - angled triangle as well.

7 0
3 years ago
Prove that the value of the expression<br> (16^3–8^3)(4^3 +2^3) is divisible by 63.
Darya [45]

Answer:

Step-by-step explanation:

16^3 - 8^3

Take out 8^3 as a common factor.

16^3 = 2^3 * 8^3

8^3(2^3 - 1)  

8^3(8 - 1) = 8^3 * 7

(4^3 + 2^3)     Expand

64 + 8            Combine

72

72 = 9 * 8

Conclusion

(8*3 * 7 )(9*8)

7*9 = 63

So any number containing 63 will divide into the reduced form of

(16^3–8^3)(4^3 +2^3)

7 0
2 years ago
Allen spreads mulch to make money over the summer. At the beginning of the summer, he has $34. The table shows the total amount
Naddik [55]
The correct answer is y=1.5 x + 34 
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