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4 years ago

Simplify 5m-2(6m-5)+2a. -7m-8b. -7m+12c. 17m+12d. 17m-8

Mathematics

1 answer:

fgiga [73]4 years ago

7 0

The answer is B. -7m+12.

First distribute the -2 to 6m and -5.

5m-2(6m-5)+2

5m+(-2*6m)+(-2*-5)+2

5m+-12m+10+2

After that, combine the like terms.

5m+-12m+10+2

-5m+-12m=-7m

10+2=12

The simplified expression is -7m+12.

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2) 3x + 4y = 8 2x+y=42 a) (30, -19) b) (31, -20) c) (32,-22) What is the answer

kirza4 [7]

Answer:

Step-by-step explanation:

We can solve simultaneous equations using substitution method, elimination method or graphical method. But for this purpose, we will be using the elimination method.

3x+4y=8 Equation 1

2x+y=42 Equation 2

Multiply Equation 1 by 2 and equation 2 by 3, so as to get the same coefficient for x

2(3x+4y=8)= 6x+8y=16 Equation 3

3(2x+y=42)= 6x+3y=126 Equation 4

Subtract equation 4 from 3, to eliminate x

6x-6x=0

8y-3y= 5y

16-126= -110

We now have 5y=-110

Divide both sides by 5,

y= -110/5

= -22

Substituting for y in equation 2

2x+(-22)= 42

2x= 42+22

2x=64

x= 64/2

= 32

(x, y)

(32, -22)

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3 years ago

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shusha [124]

Answer:

0.4

Step-by-step explanation:

$14x = 35 \\ \div b \: s \: y \: 14 \\ 14x \div 14 = 35 \div 14 \\ x = 0.4$

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3 years ago

What is the complex number z=-3+3i represented in polar form

rusak2 [61]

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Step-by-step explanation:

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3 years ago

A box in a supply room contains 24 compact fluorescent lightbulbs, of which 8 are rated 13-watt, 9 are rated 18-watt, and 7 are

Marrrta [24]

Answer:

a) There is 17.64% probability that exactly two of the selected bulbs are rated 23-watt.

b) There is a 8.65% probability that all three of the bulbs have the same rating.

c) There is a 12.45% probability that one bulb of each type is selected.

Step-by-step explanation:

There are 24 compact fluorescent lightbulbs in the box, of which:

8 are rated 13-watt

9 are rated 18-watt

7 are rated 23-watt

(a) What is the probability that exactly two of the selected bulbs are rated 23-watt?

There are 7 rated 23-watt among 23. There are no replacements(so the denominators in the multiplication decrease). Then can be chosen in different orders, so we have to permutate.

It is a permutation of 3(bulbs selected) with 2(23-watt) and 1(13 or 18 watt) repetitions. So

$P = p^{3}_{2,1}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = \frac{3!}{2!1!}*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 3*\frac{7}{24}*\frac{6}{23}*\frac{17}{22} = 0.1764$