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3 years ago

Which of the following are perfect squares? 2, 4, 6, 16, 20,25, x², x³, x¹⁵, x²⁰, x²⁵

Mathematics

2 answers:

coldgirl [10]3 years ago

4 0

Answer:

4, 16, 25, $x^{2}$ , $x^{20}\\$ .

Step-by-step explanation:

Gemiola [76]3 years ago

4 0

Answer:

16, 4, 25, x², x ²⁰

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What is the inverse operation of division?

astraxan [27]

The answer is D, Multiplication

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3 years ago

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arrange the circles (represented by their equations in general form) in ascending order of their radius lengths, x^+y^-2x+2y-1=

Lubov Fominskaja [6]

From the order you gave me in the question the answer is equation number
4 1 3 2 6 5
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3 years ago

Identify the equation that translates y=ln(x) five units down.

kari74 [83]

We are given equation: y=ln(x) .

And translation is 5 units down.

We know the rule of transformation f(x) = y + k, if k is greater than 0 it would move up or if k is less than 0, it would move down.

For the given translation 5 unit down, 5 should be less than 0.

Number less than 0 are negative numbers.

So, we need to add -5 in the given function to ln(x).

So, we would get the final equation for five units down as

y =ln(x) - 5.

6 0

3 years ago

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Please help me solve this

lesantik [10]

Speed = distance over time.
300/5 is the same as 300÷5, which is 60

5 0

3 years ago

What 2 numbers add to get 2 and multiply to 4

rusak2 [61]

Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4

first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x= $\frac{ -b+/-\sqrt{b^{2}-4ac} }{2a}$ so

1x^2-2x+4=0
a=1
b=-2
c=4
x= $\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}$
x= $\frac{ 2+/-\sqrt{4-16} }{2}$
x= $\frac{ 2+/-\sqrt{-12} }{2}$
we have $\sqrt{-12}$ and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x= $\frac{ 2+/-2\sqrt{-3} }{2}$
i= $\sqrt{-1}$
x= $\frac{ 2+/-2i\sqrt{3} }{2}$
x= $1+/-i\sqrt{3}$
x= $1-i\sqrt{3}$ or x= $1+i\sqrt{3}$ (those are the 2 numbers)

6 0

4 years ago