All categories

3 years ago

5×5×5×5 as an exponent

Mathematics

2 answers:

UNO [17]3 years ago

5 0

5^4 power. We can identify this because 5 is multiplying itself 4 times.

Hope this helps!

aleksley [76]3 years ago

4 0

5 to the second power is 4^4 or 5*5*5*5=625

You might be interested in

Joshua and his four friends have 1/2 pakage of a crackers. How many pakages of crackers do they have in all

maks197457 [2]

answer:

2.5 or 2 1/2

work:

4(joshuas friends)+1(joshua)=5

5×1/2=5/2

5/2= 2.5

7 0

2 years ago

A robot’s height is 1 meter 20 centimeters. How tall is the robot in millimeters?

Flura [38]

Answer:

Step-by-step explanation:

There are 1000 mm in one meter

and there are 10 mm in one cm

1000 + 20(10) = 1200mm

Therefore the answer is 1200mm

8 0

1 year ago

Read 2 more answers

Jia is skipping a rock across a pond. The sequence {4.2, 3.57, 3.0345, 2.5793, …} describes the height of the rock on each succe

mylen [45]

Answer:

$u_n=4.2(0.85)^{n-1}$

Step-by-step explanation:

$u_1=4.2\\u_2=3.57\\u_3=3.0345\\u_4=2.5793$

Geometric formula sequence: $u_n=ar^{(n-1)}$

(where $a$ is the first term of the sequence and $r$ is the common ratio)

To find the common ratio, divide one of the terms by the previous term:

$r=\frac{u_2}{u_1} =\frac{3.57}{4.2} =0.85$

From inspection, $a=4.2$

Therefore, $u_n=4.2(0.85)^{n-1}$

8 0

1 year ago

Read 2 more answers

WILL MARK BRAINLIEST PLEASE HELP!!!!!!!!!!!!!!!!!!

defon

Answer

Step-by-step explanation:

I am not an artist so

8 0

2 years ago

Find x Thanks for the help in advance!

Marina86 [1]

I'm not sure if this is the easiest way of doing this, but it surely work.

Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

$AH=2,\quad HB=6,\quad AC=x$

Moreover, let CH=y and BC=z

Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

$\begin{cases}4+y^2=x^2\\36+y^2=z^2\\x^2+z^2=64\end{cases}$

If we solve the first two equations for y squared, we have

$y^2=x^2-4\\y^2=z^2-36$

And we can deduce

$z^2 = x^2+32$

So that the third equation becomes

$x^2+x^2+32=64 \iff 2x^2 = 32 \iff x^2=16 \iff x=4$

(we can't accept the negative root because negative lengths make no sense)

7 0

2 years ago