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2 years ago

Simplify the expression (5jk)^4

2 answers:

8 0

First, you distribute for each number and variable. So, 5^4=5*5*5*5 which gets you 625. The rest is j^4 and k^4
625j^4k^4

ivanzaharov [21]2 years ago

4 0

The answer will be 625j^4 k^4

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-3x-3y=3 and y=5x-17

Vesna [10]

The answer for this question is "y = -5/3 and x = 8/3"

5 0

3 years ago

Last time, one out of every 4 students in our class was absent. This time, after Jen came back, only one out of every 5 students

VMariaS [17]

Answer: There are 20 students and 4 students are not here today.

Step-by-step explanation:

If the number of students = 4

Let the number of students be 'x'.

Fraction of students absent = $\dfrac{x}{4}$

If the number of students = 5

Fraction of students absent = $\dfrac{x}{5}$

And Jen came back,

So, the fraction of student absent is also written as

$\dfrac{x}{4}-1$

According to question, it becomes,

$\dfrac{x}{5}=\dfrac{x}{4}-1\\\\\dfrac{x}{5}=\dfrac{x-4}{4}\\\\4x=5(x-4)\\\\4x=5x-20\\\\4x-5x=-20\\\\-x=-20\\\\x=20$

Hence, there are 20 students in our school.

And number of students are not here today is $\dfrac{20}{5}=4$

3 0

3 years ago

The table below shows two equations:

CaHeK987 [17]

Answer:

The solutions to equation 1 are x = 3, −1.5, and equation 2 has no solution.

Step-by-step explanation:

Rearranging the two equations, you get ...

|4x -3| = 9 . . . . . has two solutions
|2x +3| = -5 . . . . has no solutions (an absolute value cannot be negative)

The above-listed answer is the only one that matches these solution counts.

_____

Testing the above values of x reveals they are, indeed, solutions to Equation 1.

4 0

3 years ago

Read 2 more answers

How many groups of 10 in 100?

PSYCHO15rus [73]

Answer:

10.

Step-by-step explanation:

You get this answer by doing 10 x 10, which is 100.

10 groups of 10 in 100.

3 0

3 years ago

Read 2 more answers

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0

3 years ago