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11111nata11111 [884]
3 years ago
14

15 yards to 19 yards

Mathematics
1 answer:
enyata [817]3 years ago
3 0
15×19=? figure it out
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The time t required to drive certain distance varies inversely with the speed r. It takes 2 hours to drive the distance at 40 mi
Roman55 [17]

Answer:

\frac{16}{9}  hours

Step-by-step explanation:

Whenever 2 quantities (suppose a and b) are inversely proportional to each other, we can write an proportionality equation (with k as proportional constant) as:

a=\frac{k}{b}

Since, here t is INVERSELY PROPORTIONAL to r, we can say:

t=\frac{k}{r}

Firstly, we have t = 2 hour and r = 40 mph, so we find k:

t=\frac{k}{r}\\2=\frac{k}{40}\\k=2*40\\k=80

Now, secondly, we have r = 45 and k = 80, we need to find t:

t=\frac{80}{45}\\t=\frac{16}{9}

So, it would take  \frac{16}{9}  hours

4 0
3 years ago
Alison knits 1/10 of scarf in 4/5 of an hour. what fraction of a scarf can alison knit in 1 hour? and how do u solve it?
pav-90 [236]
For this case we can make the following rule of three:
 1/10 scarf ------> 4/5 hour
 x ------------------> 1 hour
 Clearing the value of x we have:
 x = (1 / (4/5)) * (1/10)
 Rewriting we have:
 x = (5/4) * (1/10)
 x = 5/40
 x = 1/8
 Answer:
 
A fraction of a scarf that alison can knit in 1 hour is:
 
x = 1/8
8 0
3 years ago
Read 2 more answers
Which expression is it equivalent to?
horrorfan [7]
Option A) Is the answer. \boxed{\mathbf{\dfrac{3f^3}{g^2}}}

For this question; You are needed to expose yourselves to popular usages of radical rules. In this we distribute the squares as one-and-a-half fractions as the squares eliminate the square roots. So, as per the use of fraction conversion from roots. It becomes relatively easy to solve and finish the whole process more quicker than everyone else. More easier to remember.

Starting this with the equation editor interpreter for mathematical expressions, LaTeX. Use of different radical rules will be mentioned in between the steps.

Radical equation provided in this query.

\mathbf{\sqrt{\dfrac{900f^6}{100g^4}}}

Divide the numbered values of 900 and 100 by cancelling the zeroes to get "9" as the final product in the next step.

\mathbf{\sqrt{\dfrac{9f^6}{g^4}}}

Imply and demonstrate the rule of radicals. In this context we will use the radical rule for fractions in which a fraction with a denominator of variable "a" representing a number or a variable, and the denominator of variable "b" representing a number or a variable are square rooted by a value of "n" where it can be a number, variable, etc. Here, the radical of "n" is distributed into the denominator as well as the numerator. Presuming the value of variable "a" and "b" to be greater than or equal to the value of zero. So, by mathematical expression it becomes:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}, \: \: a \geq 0 \: \: \: b \geq 0}}

\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{\sqrt{g^4}}}

Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}

\mathbf{Since, \quad \sqrt{g^4} = g^{\frac{4}{2}}}

\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{g^2}}

Exhibit the radical rule for two given variables in this current step to separate the variable values into two new squares of variables "a" and "b" with a radical value of "n". Variables "a" and "b" being greater than or equal to zero.

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}, \: \: a \geq 0 \: \: \: b \geq 0}}

So, the square roots are separated into root of 9 and a root of variable of "f" raised to the value of "6".

\mathbf{\therefore \quad \dfrac{\sqrt{9} \sqrt{f^6}}{g^2}}

Just factor out the value of "3" as 3 × 3 and join them to a raised exponent as they are having are similar Base of "3", hence, powered to a value of "2".

\mathbf{\therefore \quad \dfrac{\sqrt{3^2} \sqrt{f^6}}{g^2}}

The radical value of square root is similar to that of the exponent variable term inside the rooted enclosement. That is, similar exponential values. We apply the following radical rule for these cases for a radical value of variable "n" and an exponential value of "n" with a variable that is powered to it.

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^n} = a^{\frac{n}{n}} = a}}

\mathbf{\therefore \quad \dfrac{3 \sqrt{f^6}}{g^2}}

Again, Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}

\mathbf{Since, \quad \sqrt{f^6} = f^{\frac{6}{2}} = f^3}

\boxed{\mathbf{\underline{\therefore \quad Required \: \: Answer: \dfrac{3f^3}{g^2}}}}

Hope it helps.
8 0
3 years ago
Which of the pairs of events below is mutually exclusive?
Nady [450]

Answer:

Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards.

Step-by-step explanation:

Given four pair of events:

1. Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards

2. Drawing a 2 and drawing a 4 with replacement from a standard deck of cards

3. Drawing a heart and then drawing a spade without replacement from a standard deck of cards

4. Drawing a jack and then drawing a 7 without replacement from a standard deck of cards

To find:

Which of the pairs is mutually exclusive?

Solution:

First of all, let us have a look at the definition of mutually exclusive events.

Mutually exclusive events do not have any case in common.

In other words, two events are known as mutually exclusive when both the events can not occur at the same time.

Now, let us have a look at the given options one by one:

1. Drawing an ace of spades and then drawing another ace of spades without replacement from a standard deck of cards.

We know that there is only one ace of spades in a standard deck.

Therefore, the two events can not occur one after the other, so these are mutually exclusive events.

2. Drawing a 2 and drawing a 4 with replacement from a standard deck of cards.

The two events can occur one after the other, so these are not mutually exclusive.

3. Drawing a heart and then drawing a spade without replacement from a standard deck of cards.

Separate cards are drawn one after the other, so not mutually exclusive.

4. Drawing a jack and then drawing a 7 without replacement from a standard deck of cards.

Separate cards are drawn one after the other, so not mutually exclusive.

3 0
3 years ago
Pls help asap! Due in 5 minutes!!! Pls
skelet666 [1.2K]

Answer: The value of X is 26.

Explanation: The explanation is in the image.

5 0
2 years ago
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