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zloy xaker [14]
3 years ago
12

Five and one third minus three and five sixths

Mathematics
2 answers:
r-ruslan [8.4K]3 years ago
6 0

Answer:

1 1/2

Step-by-step explanation:

Anni [7]3 years ago
3 0
Hello!
I’m pretty sure the answer is 1 1/2 because...
5-3=2
So we’re left with:
1/3-2/5
The Lcd of both fractions are
2/5-5/6
That equals
-3/6
When reducing the fraction
-3/6=-1/2
So combining
2-1/2
The answer is 1 1/2 or One and One half :)
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Help please i really need help
Norma-Jean [14]

The change in the stock market from the beginning of the day to the end of the day is 29 3/4

<h3>How to determine the change in the stock market from the beginning of the day to the end of the day?</h3>

The given parameters are:

Beginning = 60 3/4

End = 90 1/2

The change in the stock market from the beginning of the day to the end of the day is calculated as;

Change = End  - Beginning

So, we have

Change = 90 1/2 - 60 3/4

Evaluate

Change = 29 3/4

Hence, the change in the stock market from the beginning of the day to the end of the day is 29 3/4

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7 0
2 years ago
Solve for y.<br> 7y - 6y - 10 = 13
Orlov [11]

Answer:

y= 23

Step-by-step explanation:

Solve for  (y )

by simplifying both sides of the equation, then isolating the variable.

7y - 6y - 10 = 13

7 0
3 years ago
In the unit circle, the length of arc intercepted by an angle equals the radian measure of this angle. True False
Softa [21]
True. would be the righh answer
4 0
3 years ago
How do u do b??????????
horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

a)

if she uses and input of 6 on the 2nd one, the result will be 6² - 6  = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

b)

clearly we can never get an output  of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5

5 0
3 years ago
Betty wants to bake the perfect cookie. She wants to determine the ideal temperature at which to bake the cookies. She has enoug
borishaifa [10]

The true option about the randomized design is (c) Betty should prepare all 60 cookies.

Each cookie will receive a number.

The numbers will be written on slips of paper, put into a hat, and shuffled well.

The first 15 numbers drawn will indicate the cookies to be baked at 300°, the next 15 will indicate the cookies to be baked at 325°, and so on until all 60 cookies are allocated.

The given parameters are:

n=60

batches=4

cookies=15 ---- the number of cookies in each batch

Temperature=300,325,350,375.

To randomly assign the cookies to the given temperatures, all cookies have to be subjected to the same conditions (except for the difference in the temperature).

<h3>What is the temperature?</h3>

Normal human body temperature is the typical temperature range found in humans. The normal human body temperature range is typically stated as 36.5–37 °C. Human body temperature varies.

This means that she can bake all the cookies at once at different temperatures while noting down the cookies with a given number.

Hence, the true option is (c)

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6 0
2 years ago
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