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tangare [24]
3 years ago
14

The distance between the front wheels of a model car is 4.5 centimeters. What is the actual distance on the car if the scale is

Mathematics
1 answer:
Kamila [148]3 years ago
3 0
Answer:

108

Steps:

24 x 4.5 = 108
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Reflect the given triangle over<br> the y-axis.<br> 3 6 31<br> -3 3 3
dedylja [7]

Answer:(-3,6)(-31,-3)(-3,3)

Step-by-step explanation:

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3 years ago
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For all values of x, f(x)=2x^2 and g(x)= x+1<br> Solve fg(x) = gf(x)
Andrej [43]

Answer:

-0.25

Step-by-step explanation:

→ Find fg(x)

2 ( x + 1 )² = 2x² + 4x + 2

→ Find gf(x)

2x² + 1

→ Equate them

2x² + 1 = 2x² + 4x+ 2

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4 0
2 years ago
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What is 20% of 500ml
Nadusha1986 [10]
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8 0
4 years ago
What's the answer for this question? (The numbers after the letters are indexes btw) 27a9 x 18b5 x 4c2 Over 18a4 x 12b2 x 2c
zysi [14]

Answer:

\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c

Step-by-step explanation:

Given

\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}

Required

Simplify

\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c}

Cancel out 18

\frac{27a^9 * b^5 * 4c^2 }{a^4 * 12b^2 * 2c}

Divide 4 and 2

\frac{27a^9 * b^5 * 2c^2 }{a^4 * 12b^2 *c}

Divide 27 and 12 by 3

\frac{9a^9 * b^5 * 2c^2 }{a^4 * 4b^2 *c}

Apply law of indices

\frac{9a^{9-4} * b^{5-2} * 2c^{2-1} }{4}

\frac{9a^5 * b^3 * 2c }{4}

Divide 2 and 4

\frac{9a^5 * b^3 * c}{2}

\frac{9a^5b^3c}{2}

Rewrite as:

\frac{9}{2}a^5b^3c

Hence:

\frac{27a^9 * 18b^5 * 4c^2 }{18a^4 * 12b^2 * 2c} = \frac{9}{2}a^5b^3c

4 0
3 years ago
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