Model the number 5.55 on the place value chart
1 answer:
Answer:
Here's what I get
Step-by-step explanation:
Each digit in a number has a place value, based on its distance from the decimal point.
The value of a digit is the digit multiplied by its place value.
The place value of a digit increases by a factor of 10 as we move left on the place value chart and decreases by a factor of 10 as we move right.
Numbers to the left of the decimal point increase in the order: ones, tens, hundreds, thousands, etc.
Numbers to the right of the decimal point decrease in the order: tenths, hundredths, thousandths, etc.
So, we put the digits of 5.55 on the chart according to their place value.
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Answer:
errr polynomials ahaha
so basically we can multiply the power outside of the parantheses with the exponents on the inside:
so
this will give us:
-3x^8y^4
we also unfortunately, have to use the power on the coefficient -3 as well and we cant just MULTIPLY -3 by 4, we are going to have to multiply -3 with -3 4 TIMES(anytime we see a coefficient we MUST multiply the coefficient by itself according to the number outside of the parantheses)
so:
-3*-3*-3*-3=81
final expression
Answer:
The Line integral is π/2.
Step-by-step explanation:
We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:
we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence
(Remember that c is treated like a constant just for the x-variable).
This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.
Where, again, the constant of integration depends on Z.
As a result,
if we derivate f over z, we obtain
That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)
The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.