HOw dO I rUn??????????????????????????????????????????????

Mathematics

2 answers:

Semmy [17]3 years ago

7 0

How do you run???????

ZanzabumX [31]3 years ago

7 0

It's like walking but faster, so you go faster like sonic lol.

You might be interested in

How do you solve 3/4 (5+x) + 4 = 3x + 1? Thanks:)

elena-s [515]

$\dfrac{3}{4}(5+x)+4=3x+1\ \ \ \ |\cdot4\\\\3(5+x)+16=12x+4\ \ \ |\text{use distributive property}\\\\(3)(5)+(3)(x)+16=12x+4\\\\15+3x+16=12x+4\\\\31+3x=12x+4\ \ \ \ |-31\\\\3x=12x-27\ \ \ \ |-12x\\\\-9x=-27\ \ \ \ |:(-9)\\\\\boxed{x=3}$

3 0

3 years ago

For two years, two samples of fish were taken from a

MissTica

Answer:

Step-by-step explanation:

Trout increased its predicted average population.

6 0

3 years ago

Which best describes the slope of the given line?

dolphi86 [110]

Answer:

undefined

Step-by-step explanation:

The line shown is a vertical line

A vertical line has a slope of undefined

8 0

3 years ago

Simplify the following Questions. (IMPORTANT NOTE!!! #'s 2 AND 7 MUST BE IN SCIENTIFIC NOTATION!!!!!) 1.) (x)²(2xy³)^5 2.) (4x10

poizon [28]

Answer:

Step-by-step explanation:

Hint :

$(a*b)^{m}= a^{m}b^{m}\\\\\\(a^{m})^{n}=a^{mn}\\\\\\a^{m}*a^{n}=a^{m+n}\\$

$1)x^{2}(2xy^{3})^{5}=x^{2}*2^{5}*x^{5}*(y^{3})^{5}\\\\\\=x^{2}*2^{5}*x^{5}*y^{3*5}\\\\=x^{2}*2^{5}*x^{5}*y^{15}\\\\=2^{5}*x^{2+5}*y^{15}\\\\=2^{5}x^{7}y^{15}$

$2) (4x10^{8})^{2}=4^{2}*x^{2}*(10^{8})^{2}\\\\=16*x^{2}*10^{8*2}\\\\=1.6*10*x^{2}*10^{16}\\\\=1.6x^{2}*10^{16+1}\\\\=1.6x^{2}10^{17}$

$3)(-h^{4})^{5}=(-h)^{4*5}=-h^{20}\\\\\4)(3xy^{3})^{2}(xy)^{6}=3^{2}x^{2}(y^{3})^{2}x^{6}y^{6}\\\\=9x^{2}y^{3*2}x^{6}y^{6}\\\\\\=9x^{2}y^{6}x^{6}y^{6}\\\\=9x^{2+6}y^{6+6}\\\\=9x^{8}y{12}\\\\\\$

$5) (p^{9})^{-2}=p^{9*-2}=p^{-18}\\\\\\6)(5k^{2})^{3}=5^{3}(k^{2})^{3}=625k^{6}\\\\7)(7x10^{5})^{2}=7^{2}x^{2}(10^{5})^{2}\\\\=49x^{2}10^{5*2}\\\\=4.9*10^{1}x^{2}10^{10}\\\\=4.9x^{2}10^{10+1}\\\\=4.9x^{2}10^{11}$

$8)x^{3}(-x^{3}y)^{2}=x^{3}*(-x^3})^{2}y^{2}\\\\\\=x^{3}*(-1)^{2}*x^{2*3}y^{2}\\\\=x^{3}*1*x^{6}y^{2}=x^{3+6}y^{2}\\\\=x^{9}y^{2}\\\\\\9)w^{5}(w^{2})^{-4}=w^{5}w^{2*-4}\\\\=w^{5}w^{-8}\\\\=w^{5-8}=w^{-3}\\\\=\frac{1}{w^{3}}\\\\\\10) (2x^{5})^{4}=2^{4}x^{5*4}\\\\=2^{4}x^{20}\\\\=16x^{20}$