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3 years ago

Fill in the blanks to explain how to solve 3/5 divided by 1/10.

1 answer:

7 0

You can use a common denominator and rewrite 3/5 divided by 1/10 as 6/10 divided by 1/10.
Than you can think, “How many ___s are in ___? (Sorry, i don’t know this part)
3/5 divided 1/10 is 6

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Please help me with this

Vera_Pavlovna [14]

Answer:

226.19

Step-by-step explanation:

pi * r^2 = area

The radius = diameter/2 = 6

pi * 6^2 * 2 = area of 2 tortillas

72pi = area of 2 tortillas

That rounds to 226.19

3 0

3 years ago

Given the function g(x) = 6(4)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Natasha2012 [34]

I also need this answer so when you find out let me know

5 0

3 years ago

Find the length of side X in simplest radical form with a rational denominator.<br> X

german

Answer:

given in the figure an isosceles right triangle

hence

x = √3

√3 is a <u>surd</u> it can't be simplified further

7 0

3 years ago

Two dice are rolled, and the sum of the face values is six. What is the probability that at least one of the dice came up a thre

gizmo_the_mogwai [7]

Answer : $P(\text{ atleast one of the dice come up 3})= \frac{1}{5}$

Explanation :

We know that ,

Sample space of " sum of faces is 6"= $\{(3,3),(4,2),(2,4),(5,1),(1,5)\}$

Atleast one of the dice come up 3 = $\{(3,3)\}$

So,

$Probability = \frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}}$

$P(\text{ atleast one of the dice come up 3})= \frac{1}{5}$

8 0

4 years ago

Step 3: Now multiply the values to get 14,918,904,000. That is almost 15 billion passwords.

Troyanec [42]

The number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600

<h3>How to determine the number of passwords</h3>

The given parameters are:

Password length = 8

Letters = 6

Numbers = 2

There are 26 letters and 10 digits.

Since the letters and the numbers can be repeated, then the number of passwords is:

Count = 26^6 * 10^2

Evaluate

Count = 30891577600

Hence, the number of passwords that would be allowed if users were allowed to reuse the letters and numbers is 30891577600