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LUCKY_DIMON [66]
2 years ago
12

Craig Campanella and Edie Hall both have night jobs. Craig has every thirteenth night and Edie has every Fifth night off. How of

ten will they have same night off?
Mathematics
1 answer:
Andrews [41]2 years ago
6 0

Craig has every 13th night and Edie has every 5th night off.

You have to find LCM - the Least Common Multiple that is the smallest ("least") number that both 13 and 5 will divide into.

Since numbers 13 and 5 are both prime, then LCM(13,5)=13·5=65.

This means, they will have the same every 65th night off.  

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Ellie completed an algebraic proof to show that V12 + V108 = 8 - 31.
lorasvet [3.4K]

Answer:

The answer to your question is given below.

Step-by-step explanation:

To know the correct answer to the question, do the following:

√12 + √108

= 12^½ + 108^½

= 4^½ • 3^½ + 36^½ • 3^½

Factorise

3^½ (4^½ + 36^½)

Recall:

4^½ = √4 = 2

36^½ = √36 = 6

Therefore,

3^½ (4^½ + 36^½) = 3^½ (2 + 6)

= 3^½ • 8

8 0
2 years ago
The statement below describes a situation in which opposite quantities combine to make 0. Is the statement true or false? A plan
Nadusha1986 [10]
I believe this is true....lets say the plane takes off and goes up 500 ft.....it circles, but then it has to come back down...land...so it comes down 500 ft...
500 - 500 = 0
5 0
2 years ago
Read 2 more answers
2.8, 3.4, 4.0, 4.6, . . . Write an equation for the nth term of the arithmetic sequence. Then find a50.
vova2212 [387]

Answer:

a_n = 2.2 + 0.6 n

a_50 = 32.2

Step-by-step explanation:

What's the common difference of this series?

a_1 = 2.8

a_2 = 3.4

Common difference = a_2 - a_1 = 3.4 - 2.8 = 0.6.

Expression for the nth term:

a_n = a_1 + (n - 1) \cdot\text{Common Difference} \\\phantom{a_n } = 2.8 + 0.6 \; (n-1) \\\phantom{a_n} = 2.8 + 0.6 \; n - 0.6\\\phantom{a_n} = 2.2 + 0.6\; n

n = 50 for the fiftieth term. Therefore

a_{50} = 2.2 + 0.6 \times 50 = 32.2.

6 0
2 years ago
What is the product? StartFraction x squared minus 16 Over 2 x + 8 EndFraction times StartFraction x cubed minus 2 x squared + x
zubka84 [21]

Answer:

[x(x - 1)(x - 4)]/(2(x + 4))

Step-by-step explanation:

We want to find;

[(x² - 16)/(2x + 8)] * [(x³ - 2x² + x)/(x² + 3x - 4)]

Now,

x² - 16 can be factorized as;

(x + 4)(x - 4)

Also, 2x + 8 can be factorized as;

2(x + 4)

Also, (x³ - 2x² + x) can factorized as;

x[x² - 2x + 1] = x[(x - 1)(x - 1)]

Also,(x² + 3x - 4) can be factorized out as; (x - 1)(x + 4)

So plugging in these factorized forms into the equation in the question, we have;

[(x + 4)(x - 4)/(2(x + 4))] * [x[(x - 1)(x - 1)] /((x - 1)(x + 4))

This gives;

((x - 4)/2) * x(x - 1)/(x +4)

This gives;

[x(x - 1)(x - 4)]/(2(x + 4))

8 0
2 years ago
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EVELYN HAD 9 DOLLS, BUT SHE LOST 1/3RD OF THEM AT DAYCARE. HOW MANY TOTAL DOLLS WOULD EVELYN HAVE HAD IF SHE HAD NOT LOST THEM?
Elanso [62]

Answer:

Total Dolls would Evelyn have had if she had not lost them = 9 Dolls

Step-by-step explanation:

As given,

Total dolls Evelyn had = 9

Total dolls lost = \frac{1}{3}× 9 = 3

So, Now

Evelyn had total dools after lost = 9 - 3 = 6

If she had not lost te dolls , then she had 3 dolls more

∴ we get

If she had not lost any dolls , Evelyn had total dolls = 6 + 3 = 9

So, The answer would be :

Total Dolls would Evelyn have had if she had not lost them = 9 Dolls

3 0
1 year ago
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