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3 years ago

13(6x – 5) – x = 13 – 2(x + 1

Mathematics

1 answer:

elena-s [515]3 years ago

6 0

Answer:

Step-by-step explanation:

$13(6x-5)-x=13-2(x+1)\\78x-65-x=13-2x-2\\77x-65=11-2x\\77x+2x=11+65\\x=76/79\\x=0.9620$

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Least common multiple <br> 14,6

Vlada [557]

Answer:

Find the prime factorization of 14

14 = 2 × 7

Find the prime factorization of 6

6 = 2 × 3

Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:

LCM = 2 × 3 × 7

LCM = 42

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3 years ago

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3 feet of ribbons are needed for wrapping each present. How many presents can be wrapped with 102cm of ribbons?

bagirrra123 [75]

1 imperial foot has approximately 30.48 metric centimeters, and for one present we need 3 ft, or namely 3(30.48) cm, how many can we get from 102 cm? 102 ÷ 3(30.48) ≈ 1.1155, so barely just one present.

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maks197457 [2]

Answer:

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Step-by-step explanation:

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$\frac{4 \\boys}{12 \\ \\girls}$

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3. In the original equation, the amount of girls is 3 times the amount of boys

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3 years ago

Use law of sines or law of cosines to find the length of side AB Please show work !!

scoray [572]

Answer:

$AB = 11.3$

Step-by-step explanation:

Given

The attached triangle

Required

Length AB

To do this, we make use of sine law which is represented as:

$\frac{a}{\sin(a)} = \frac{b}{\sin(b)} = \frac{c}{\sin(c)}$

So, we have:

$\frac{15}{\sin(70)} = \frac{AB}{\sin(45)}$

Make AB the subject

$AB = \frac{15}{\sin(70)} * \sin(45)$

$AB = \frac{15}{0.9397} *0.7071$

$AB = 11.3$

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3 years ago