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stellarik [79]
3 years ago
7

B) Sajid makes a scale model of a lorry.

Mathematics
1 answer:
padilas [110]3 years ago
3 0
If Sajid’s scale is 1:32, the ratio of the length of his model and the lorry is, of course, 1/32. Meaning that the lorry is (32*45)cm long. So, it is 1440 cm. If Chitra’s model uses the scale 1:72, every 72 cm of length of the lorry is 1 cm on the model, meaning that the length of his model is (1440/72) cm= 20 cm
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Given: △ABC, m∠B=90° AB=12, BC=16, BK ⊥ AC . Find: AC and BK. Please explain how to find out BK​
Kryger [21]

Answer:

Given: △ABC, m∠B=90° AB=12, BC=16, BK ⊥ AC . Find: AC and BK.

Given: △ABC, m∠B=90°

Find: AC and BK.

Short leg 90 degrees Long leg Hypotenuse

AB=12 90 BC=16 AC= ?

AK = ? 90 BK = ? AB=12

AC = SQRT (AB*AB + BC*BC) = 20 [right triangle; Pythagorean Theorem]

Similar triangles:[Note: In diagram, share two angles. Therefore share three angles]

BK / 16 = AB / AC

BK / 16 = 12 / 20

BK = (3/5)16

BK = 48/5

another answer let see this

AB^2+BC^2=AC^2

12^2+16^2=AC^2

144+256=AC^2

400=AC^2

20=AC

# be careful#

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2 years ago
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WINSTONCH [101]
The answer is 6 triangles, one hexagon. Hope this helps
5 0
3 years ago
List from least to greatest 0,-5.5,-15,15,25,-25
S_A_V [24]

Answer:

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Step-by-step explanation:

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Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract
anyanavicka [17]

Answer:

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8

Step-by-step explanation:

The sum you are trying to understand is this.

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}

Remember that in general when you have a geometric series  

\sum\limits_{n = 0}^{\infty} a*r^n you have that

\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}      and that equality is true as long as     |r| < 1.

Therefore here we have

\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n        and   \big|\frac{-1}{15} \big| = \frac{1}{15} < 1

Therefore we can use the formula and

\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n =  \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16  = 15/8

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Original: 12 for $9.12<br> New Deal: 9 for
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Answer:

6.84

Step-by-step explanation:

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