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ryzh [129]
4 years ago
10

Can someone help me with this question please. I'll give the brainliest answer to whoever helps me.

Mathematics
1 answer:
Doss [256]4 years ago
5 0

Answer: \sqrt[5]{y}

I realize its probably not the largest readable font. If you are having trouble reading it, it is the square root of y; however, there is a tiny little 5 in the upper left corner to indicate a fifth root. So you would read it out as "the fifth root of y"

The rule I'm using is

x^{1/n} = \sqrt[n]{x}

and the more general rule we could use is

x^{m/n} = \sqrt[n]{x^m}

where m = 1. This rule helps convert from rational exponent form (aka fractional exponents) to radical form.

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A fair eight-sided die with the sides numbered 1 to 8 is rolled twice.
ryzh [129]

A - rolling an odd prime number followed by an even prime number

B - two numbers are consecutive numbers in ascending order


|\Omega|=8^2=64\\\\ |A|=\underbrace{3}_{\text{3,5,7}}\cdot\underbrace{1}_{\text{2}}=3\\ |B|=\underbrace{7}_{\text{(1,2),(2,3),\ldots(7,8)}}\\\\\\ P(A)=\dfrac{3}{64}\approx5\%\\ P(B)=\dfrac{7}{64}\approx11\%

4 0
4 years ago
Write 1.4 over 7 in simplest form
Nana76 [90]
2/10 because 0.7/7 is equal to one tenth, so 1.4/7 is equal to 2/10
7 0
3 years ago
7/5= blank write improper fraction as a mixed number
Ostrovityanka [42]
Hello!

To do this you see how many times 5 goes into 7. As you can see it only goes in once. This gives us a whole number one. Since we took away 5 we have two sevens left. This gives us the mixed number below.

1 \frac{2}{7}

I hope this helps!
8 0
3 years ago
What is the approximate area of the circle shown below?
svet-max [94.6K]

Answer:

A. 804 cm²

Step-by-step explanation:

Formula :

\boxed{A = \pi r^2}

A = 3.14 × 16²

A = 3.14 × (16 × 16)

A = 3.14 × 256

A = 803.84 cm²

A = <u>8</u><u>0</u><u>4</u><u> </u><u>cm²</u> (rounded)

So, the approximate area of the shown below is 804 cm² (A)

#CMIIW ^^

3 0
4 years ago
Given six consecutive integers with a sum of seven times the
Furkat [3]

Answer:

<em>x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 7(x + 1)</em>

Step-by-step explanation:

<u>Equations</u>

The situation can be written as an algebraic equation by setting the variables as follows:

x = first integer

x + 1 = second integer

x + 2 = third integer

x + 3 = fourth integer

x + 4 = fifth integer

x + 5 = sixth (and last) integer

The sum of all six numbers must be equal to 7 times the second number, thus:

x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 7(x + 1)

6 0
3 years ago
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