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Nostrana [21]
2 years ago
12

21x - 56 = -2 what is x?

Mathematics
2 answers:
kaheart [24]2 years ago
6 0

Answer:

I got 2.5

Step-by-step explanation

1) add 56 to both sides to get 21x = 54

2) divide both sides by 21

3) you should get 2.5

den301095 [7]2 years ago
5 0

Answer:

About 2.57

Step-by-step explanation:

21x-56=-2 \\\\21x=54 \\\\x\approx 2.57

Hope this helps!

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Chelsey put for $450 into an account with a simple interest rate of 2.5% when she withdrew the money she had earned a total of $
fgiga [73]

Answer:

5.66 years

Step-by-step explanation:

450 x 1.025^(n) = 450 + 67.50

450 x 1.025^(n) = 517.50

1.025^(n) =(517.50 / 450)

1.025^(n) = 1.15

n = ln(1.15) / ln(1.025)

n = 5.66

6 0
2 years ago
A box in a supply room contains 18 compact fluorescent lightbulbs, of which 6 are rated 13-watt, 8 are rated 18-watt, and 4 are
V125BC [204]

Answer:a) Probability P(exactly 2 bulbs rated 13watts)= 0.22

b) Probability(each bulb different rating)

= 0.24

Step-by-step explanation:

There are 6 13watts bulbs

8 18watts bulbs

4 23watts bulbs

Total bulbs = 18

a) Probability that all 3 bulbs are 18watts

Number of ways of pulling 3 bulbs = 18!/(3!×15!) = (6.4×10^15)/7.846×10^12) = 815 ways

Different ways of pulling 13watts bulbs out of 6 = 6!/(2!×4!)= 720/48=15ways

Different ways of pulling non 13 watts bulbs= 12!/(1!×11!) = 479,001,600/ 39,916,800 = 12

Number of ways total= 15×12=180waya

Therefore P(exactly 2 bulbs rated 13watts)= 180/815 =0.22

b) Probability P( all 3 bulbs are 1 from each rating)

Ways of pulling 3bulbs bulb each from 3 ratings are 4× 5 × 8= 192ways

Probability = 192/815 =0.24

6 0
2 years ago
Read 2 more answers
Henry is making a recipe for biscuits. A recipe calls for 5/10 kilogram flour and 9/100 kilogram sugar .
Alchen [17]
59/100 or 0.59.
Hope I helped!
~ Zoe
6 0
2 years ago
Khalil is 1.65 meters tall. At 3 p.m., he measures the length of a tree's shadow to be
igor_vitrenko [27]

Answer: 1.65

Step-by-step explanation:

4 0
1 year ago
Will crown the correct answers brainliest 3x
Solnce55 [7]

Answer :

(1) \frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}

(2) \frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{15xy}

(3) \frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{14xy}

(4) \frac{2}{5x}+\frac{3}{7y}=\frac{6y+15x}{21xy}

(5) \frac{7}{11x}-\frac{1}{33y}=\frac{231y-11x}{363xy}

(6) \frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{7y+42y-2x}{14xy}

Step-by-step explanation :

(1) The given expression is: \frac{1}{x}+\frac{1}{y}

\frac{1}{x}+\frac{1}{y}=\frac{y+x}{xy}

(2) The given expression is: \frac{1}{5x}+\frac{1}{3y}

\frac{1}{5x}+\frac{1}{3y}=\frac{3y+5x}{(5x)\times (3y)}=\frac{3y+5x}{15xy}

(3) The given expression is: \frac{1}{7x}-\frac{1}{2y}

\frac{1}{7x}-\frac{1}{2y}=\frac{2y-7x}{(7x)\times (2y)}=\frac{2y-7x}{14xy}

(4) The given expression is: \frac{2}{5x}+\frac{3}{7y}

\frac{2}{5x}+\frac{3}{7y}=\frac{(2\times 3y)+(3\times 5x)}{(5x)\times (7y)}=\frac{6y+15x}{21xy}

(5) The given expression is: \frac{7}{11x}-\frac{1}{33y}

\frac{7}{11x}-\frac{1}{33y}=\frac{(7\times 33y)-(1\times 11x)}{(11x)\times (33y)}=\frac{231y-11x}{363xy}

(6) The given expression is: \frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}

\frac{1}{2x}+\frac{3}{x}-\frac{1}{7y}=\frac{(x\times 7y)+(3\times 2x\times 7y)-(2x\times x)}{(2x)\times (x)\times (7y)}=\frac{7xy+42xy-2x^2}{14x^2y}=\frac{7y+42y-2x}{14xy}

7 0
2 years ago
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