All categories

2 years ago

Divide 4^2 divided by 4^6 =

Mathematics

2 answers:

frutty [35]2 years ago

5 0

Answer:

Answer in exponent form ⇒ 4⁻⁴
Answer in fraction form ⇒ 1/256

Step-by-step explanation:

Rule: If the bases of the term is the same when dividing, then you can simply subtract the exponents.

4² ÷ 4⁶
=> 4² ⁻ ⁶
=> 4⁻⁴
=> 1/4⁴
=> 1/256

Conclusion:

Hence,

Answer in exponent form ⇒ 4⁻⁴
Answer in fraction form ⇒ 1/256

Hoped this helped.

$BrainiacUser1357$

alekssr [168]2 years ago

3 0

Answer:

Step-by-step explanation:

so you are saying do:

$4^{2} /4^{6}$ which is 16 /4096

16/4096=0.00390625

another way you can do this is

$4^{2} /4^{6}$

just subtract the powers

2-6=-4

so the answer is $4^{-4}$

You might be interested in

is g(x)=-(x+1)^2+3 a quadratic function or exponential function??

Morgarella [4.7K]

It’s a quadratic since there is ^2 term on the x. Exponential is when you have x in the exponent like 5^x or e^x

7 0

2 years ago

a triangle has two sides of length 5 and 4 what is the smallest possible whole-number length for the third side

nikdorinn [45]

Answer:

Step-by-step explanation:

5 0

3 years ago

Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l

SIZIF [17.4K]

Answer:

$\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}$

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let

$C_1,C_2$

be the two paths.

Recall that if we parametrize a path C as $(r_1(t),r_2(t),r_3(t))$ with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt$

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and

$\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}$

The parametrization of the line segment from (2,2,2) to

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)

r'(t) = (-11,4,3)

and

$\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}$

Hence

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}$

8 0

2 years ago

Whats the answer to this problems

Airida [17]

Here's your answer to the first problem

7 0

3 years ago

Uhhhhhhhhh I need hellpppppppppp

GenaCL600 [577]

Bsuskshdjskshsosshdkshsksbxksmsjs

6 0

3 years ago